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2.
Circle
PART - I : CIRCLE - TANGENT
THEOREM
A tangent at any point of a circle is perpendicular to the radius, through
the point of contact.
In adjoining figure,
line PQ is a tangent at A and
O
Seg AO is the radius through
the point of contact A.
 seg OA  line PQ.
A
P
Q
EXERCISE - 2.1 (TEXT BOOK PAGE NO.50)
2.
In the adjoining figure,
point A is the centre of the circle.
AN = 10 cm. Line NM is tangent at M.
Determine the radius of the circle
if MN = 5 cm.
(2 marks)
Sol.





In AMN,
m AMN = 90º
AN² = AM² + MN²
10² = AM² + 5²
100 = AM² + 25
AM² = 100 – 25
AM² = 75
 AM =
75
 AM =
25  3
A
M
N
[Radius is perpendicular to the tangent]
[By Pythagoras theorem]
[Given]
 AM = 5 3 cm.
 Radius of the circle is 5 3 cm.
EXERCISE - 2.1 (TEXT BOOK PAGE NO.51)
3.
Sol.
In the adjoining figure,
M
Q is the centre of circle and PM
and PN are tangent segments to
P
40º
the circle. If MPN = 40º circle,
Q
find MQN.
(2 marks)
In MQNP,
N
m MPN = 40º
[Given]
m PMQ = 90º
[Radius is perpendicular to the tangent]
m PNQ = 90º
 m MQN = 140º
74
[Remaining angle]
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EXERCISE - 2.1 (TEXT BOOK PAGE NO.51)
4.
As shown the adjoining figure,
two concentric circles are given
and line AB is tangent to the smaller
O
circle at T. Show that T is the
(2 marks)
midpoint of seg AB.
Construction : Let O be the centre of the
B
A
T
concentric circles and draw seg OT.
Proof :
Line AB is tangent to the smaller circle at T.
[Given]
 seg OT  line AB
.......(i) [Radius is perpendicular to the tangent]
With respect to the bigger circle,
[From (i)]
seg OT  chord AB
 AT = BT
[Perpendicular from the centre of a
circle to a chord bisects the chord]
 T is the mid-point of seg AB.
EXERCISE - 2.1 (TEXT BOOK PAGE NO.50)
1.
Sol.
In the adjoining figure,
point P is centre of the circle
P
and line AB is the tangent to the
circle at T. The radius of the circle
is 6 cm. Find PB ifTPB = 60º.
(2 marks)
T
A
B
In PTB,
[Given]
m TPB = 60º
m PTB = 90º
[Radius is perpendicular to tangent]
 m PBT = 30º
[Remaining angle]
 PTB is a 30º - 60º - 90º triangle
 By 30º - 60º - 90º triangle theorem
1
PB
[Side opposite to 30º]
PT =
2
1
 6 =
PB
[Given]
2
 PB = 6 × 2
 PB = 12 cm
EXERCISE - 2.1 (TEXT BOOK PAGE NO.51)
B
In the adjoining figure,
A
line AB is tangent to both the
M
circles touching at A and B.
P
OA = 29, BP = 18, OP = 61
1
6
O
then find AB.
(3 marks)
Construction : Draw seg PM  seg OA,
A-M-O
Sol.
In PBAM,
m PBA = 90º
[Radius is perpendicular to the tangent]
m BAM = 90º
m PMA = 90º
[Construction]
 m MPB = 90º
[Remaining angle]
29
18
6.
S C H O O L S E C TI O N
75
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 PBAM is a rectangle
 PB = AM = 18 units
OA = OM + AM
 29 = OM + 18
 OM = 29 – 18
 OM = 11 units






In PMO,
m PMO = 90º
OP 2 = OM2 + PM2
(61)2= (11)2 + PM2
3721= 121 + PM2
PM2 = 3721 – 121
PM2 = 3600
PM = 60 units
But, AB = PM
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[By definition]
[Opposite sides of a rectangle]
[ A - M - O]
[Construction]
[By Pythagoras theorem]
[Opposite sides of a rectangle]
 AB = 60 units
PROBLEM SET - 2 (TEXT BOOK PAGE NO.192)
1.
Sol.
In the adjoining figure, in a circle with centre P,
a chord AB is parallel to a tangent and intersects
P
the radius drawn from a point of contact at
the midpoint of the radius. If AB = 12,
A
E
find the radius of the circle. (3 marks)
•
F
C
Let the radius of the circle be 2x units
 PA = PF = 2x units .......(i)
[Radii of same circle]
E is the midpoint of seg PF
[Given]






PE =
•
D
1
PF
2
1
 2x
2
PE = x units
line AB || line
On transversal
PEA  PFC
But, m PFC =
m PEA = 90º
seg PE  chord
PE =
AE =
B
[From (i)]
......(ii)
CD
[Given]
PF,
[Converse of corresponding angle test]
90º
[Radius is perpendicular to the tangent]
......(iii)
AB
[From (iii)]
1
× AB
2
[The perpendicular drawn from the centre
of the circle to the chord bisects the chord]


76
1
 12
2
AE = 6 units
In PEA,
m PEA = 90º
AE =
[Given]
......(iv)
[From (iii)]
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







PA2 = PE2 + AE2
[By Pythagoras theorem]
(2x)2 = x2 + 62
4x2 – x2 = 36
3x2 = 36
x2 = 12
x= 4×3
x= 2 3
[Taking square roots]
PA = 2 × 2 3 = 4 3 units

Radius of the circle is 4 3 units.
THEOREM
A line perpendicular to the radius at its outer end is a tangent to the
circle.
In adjoining figure,
O
line l is perpendicular to radius OA
at its outer end A,
 line l is a tangent.
l
A
EXERCISE - 2.1 (TEXT BOOK PAGE NO.51)
5.
Tangents to the circle with centre O, at the points A and B intersect at
P, a circle is drawn with centre P passing through A, Prove that the
tangent at A to the circle with centre P passes through O. (3 marks)
C
Proof :
Consider circle with centre O,
Line AP is a tangent to the circle at point A.
A
[Given]
 m OAP = 90º
......(i)
P
O
[ Radius is perpendicular to the tangent]
Consider circle with centre P,
B
line OA  seg AP
[From (i)]
 line OA is a tangent at point A
[A line perpendicular to the
radius at its outer end is tangent
to the circle]
 Tangent at A to the circle with centre P passes through O.

THEOREM
Statement : The lengths of the two tangent segments to a circle drawn
(4 marks)
from an external point are equal.
A
Given : (i) A circle with centre O.
(ii) P is a point in the exterior of the circle.
(iii) Points A and B are the points of contact
O
P
of the two tangents from P to the circle.
To Prove : PA = PB
B
Construction : Draw seg OA, seg OB and seg OP.
Proof : In PAO and PBO,
m PAO = m PBO = 90º[Radius is perpendicular to the tangent]
Hypotenuse OP  Hypotenuse OP [Common side]
[Radii of same circle]
seg OA  seg OB

PAO  PBO
[By hypotenuse - side theorem]
seg PA  seg PB
[c.s.c.t]

PA = PB
S C H O O L S E C TI O N
77
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EXERCISE - 2.1 (TEXT BOOK PAGE NO.51)
7.
Sol.
Explain what is wrong with the information marked on the following
C
figures (without taking actual measurements) (2 marks)
(i)
30º
3
In ABC,
5
[Given]
m A = 60º
A 60º
m C = 30º
4
m B = 90º
B
 Line BC is a tangent to the circle at point B
[A Line perpendicular to the
radius at its outer end is a tangent to the circle]
But according to given figure Line BC is not a tangent.
(ii)
O





110º
20º
D
C
Take a point D on ray AC
A
such that A - C - D
[Linear pair axiom]
m OCD + m OCA = 180º
110º + m OCA = 180º
m OCA = 180 – 110
m OCA = 70º
......(i)
[From (i)]
In OAC, m  OCA = 70º
m  COA = 20º
[Given]
m  OAC = 90º
[Remaining angle]
Line AC is a tangent
[A Line perpendicular to the
radius at its outer end is a tangent to the circle]
But according to the given figure Line AC is not a tangent.
K
13
(iii)
E







78
In OPE,
m OPE = 90º
OE2 = OP2 + EP2
152 = 92 + EP2
225 = 81 + EP2
EP2 = 225 – 81
EP2 = 144
EP = 12 Units
EP = EK = 12 Units
15
O
9
[Given]
[By Pythagoras theorem]
P
[Taking square roots]
[The lengths of two tangent segments
from an external point to a circle are equal]
But according to the given figure EK = 13 units.
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EXERCISE - 2.1 (TEXT BOOK PAGE NO.51)
8.
Sol.
In the adjoining figure,
B 3 Q
5
P
are four tangents to a circle at the
points A, B, C and D. These four tangents
form a parallelogram PQRS.
C
A
If PB = 5 and BQ = 3 then find PS. (4 marks)
PA = PB = 5
BQ = CQ = 3
R
[The lengths of the two tangent S
D
Let,
segments to a circle drawn from an
AS = SD = x
external point are equal]
CR = DR = y
PQRS is a parallelogram
[Given]

PQ = SR
[ Opposite sides of a parallelogram
are congruent]

PB + BQ = SD + DR
[P - B - Q and S - D - R]

5+3 = x+y

x+y = 8
......(i)
PS = QR
[ Opposite sides of a parallelogram
are congruent]

PA +AS = QC + CR
[ P - A - S and Q - C - R]

5+x = 3+y

x–y = 3–5

x–y = –2
......(ii)
Adding (i) and (ii)
x + y + x – y = 8 + (– 2)

2x = 8 – 2

2x = 6

x = 3
PS = PA + AS
[ P - A - S]

PS = 5 + x

PS = 5 + 3

PS = 8 units
EXERCISE 2.1 (TEXT BOOK PAGE NO.52)
12.
In the adjoining figure,
AB and AC are tangents drawn
from A, and BA  CA. Prove that
BACO is a square.
(2 marks)
Proof :
In BACO,
m OBA = 90º
m OCA = 90º
m BAC = 90º
 m BOC = 90º
 BACO is a rectangle
seg OB  seg OC
 BACO is a square
S C H O O L S E C TI O N
O
B
C
A
[Radius is perpendicular to the tangent]
[Given]
[Remaining angle]
[By definition]
[Radii of the same circle]
[A rectangle in which adjacent sides
are congruent is a square]
79
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EXERCISE - 2.1 (TEXT BOOK PAGE NO.52)
m
9.
In the adjoining figure,
point A is a common point of contact l
B
D
of two externally touching circles
C
and line l is a common tangent
to both circles touching at B and C.
Line m is another common tangent
A
at A and it intersects BC at D.
Prove that (i) BAC = 90º
(ii) Point D is the midpoint of seg BC.
(4 marks)
Proof :
In BDA,
DB = DA
.......(i)
[The lengths of the two tangent
segments from an external point
to a circle are equal]
 DBA  DAB
[Isosceles triangle theorem]
Let,
m DBA = m DAB = xº ......(ii)
In DAC,
DA = DC
......(iii)
[The lengths of the two tangent
segments from an external point
to a circle are equal]
 DAC  DCA
[Isosceles triangle theorem]
Let,
m DAC = m DCA = yº......(iv)
m BAC = m DAB + m DAC
[Angle Addition Property]
 m BAC = (x + y)º
........(v)
[From (ii) and (iv)]
In ABC,
m ABC + m ACB + m BAC = 180º
[ Sum of the measures of the
angles of a triangle is 180º]

x + y + x + y = 180 [From (ii), (iv), (v) and B - D - C]

2x + 2y = 180

2(x + y) = 180
180

x+y =
2

x + y = 90

m BAC = 90º [From (v)]
From (i) and (iii) we get,
DB = DC
 D is the midpoint of seg BC.
10.
EXERCISE - 2.1 (TEXT BOOK PAGE NO.52)
A
In the adjoining figure,
ABC is isosceles triangle with
perimeter 44 cm. The base BC
is of length 12 cm. Sides AB and
AC are congruent. A circle touches
P
Q
the three sides as shown. Find the
length of a tangent segment from
A to the circle.
(4 marks)
B
Sol.
Let, AP = AQ = x
BP = BR = y
CR = CQ = z
80
......(i)
......(ii)
......(iii)
R
C
[The lengths of the two tangent
segments to a circle drawn from
an external point are equal]
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








AB
AB
Similarly,
BC
BC
AC
Perimeter of ABC
AB + BC + AC
x+y+y+z+x+z
2x + 2y + 2z
2(x + y + z)
x+y+z
x + 12
x
x
AP

Length of a tangent segment from A to the circle is 10 cm.


=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
AP + PB
[ A - P - B]
x+y
.....(iv) [From (i) and (ii)]
y+z
12
.....(v)
x+z
.....(vi)
44 cm
[Given]
44
44
[From (iv), (v), and (vi)]
44
44
22
22
[From (v)]
22 – 12
10
AQ = 10 cm
[From (i)]
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 195)
14.
Centre of the circle is O, tangent CA
at A and tangent DB at B intersect
each other at point P. Bisectors of
CAB and DBA intersect at
Q
point Q on the circle. APB = 60º,
(4 marks)
prove that PAB QAB.
Proof :
In PAB,
PA = PB
C
•
•
A

60º
O
×
×
P
B
D
[The lengths of two tangent
segments from an external point to a circle are equal]
 PAB  PBA
.....(i) [Isosceles triangle theorem]
m PAB + m PBA + m APB = 180º
[Sum of the measures of
angles of a triangle is 180º]

m PAB + m PAB + 60º = 180º
[From (i) and given]

2 m PAB = 180 – 60

2 m PAB = 120

m PAB = 60º

m PAB = m PBA = 60º ......(ii)
[From (i)]
m PAB + m BAC = 180º
[Linear pair axiom]

60 + m BAC = 180º
[From (ii)]

m BAC = 120º
.....(iii)
Similarly,
m ABD = 120º
.....(iv)
1
m QAB = m BAC [ ray AQ bisects BAC]
2
1
=
× 120º [From (iii)]
2

m QAB = 60º
......(v)
Similarly,
m QBA = 60º
.....(vi)
In PAB and QAB,
PAB  QAB
[From (ii) and (v)]
seg AB  seg AB
[Common side]
PBA  QBA
[From (ii) and (vi)]
 PAB  QAB
[By ASA test of congruence]
S C H O O L S E C TI O N
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PROBLEM SET - 2 (TEXT BOOK PAGE NO.193)
2.
In the adjoining figure,
O is the centre and seg AB is
a diameter. At the point C on
the circle, the tangent CD is drawn.
Line BD is a tangent to the circle at
the point B. Show that seg OD || chord AC.
(4 marks)
Construction : Draw seg OC.
Proof :
In OAC,
seg OA  seg OC

OAC  OCA
Let,
m OAC = m OCA = x
In OCD and OBD,
seg OC  seg OB
seg CD  seg BD










seg OD
OCD
COD
Let,
m COD
m BOC
m BOC
m BOC
BOC is
m BOC
2y
2y
y
m BOD
BOD
seg OD
 seg OD
 OBD
 BOD
D
C
A
O
B
[Radii of same circle]
[Isosceles triangle of theorem]
........(i)
[Radii of same circle]
[The lengths of two tangent
segments from an external point
to a circle are equal]
[Common side]
[By SSS test of congruence]
[c.a.c.t]
= m BOD = y ........(ii)
= m COD + m BOD
[Angle addition property]
= y+y
[From (ii)]
= 2y
........(iii)
an exterior angle of AOC
= m OCA + m OAC
[Remote interior angles theorem]
= x+x
[From (i) and (iii)]
= 2x
= x
= m OAC
[From (i) and (ii)]
 m BAC
[A - O - B]
||chord AC
[Corresponding angles test]
PROBLEM SET - 2 (TEXT BOOK PAGE NO.193)
3.
Two circles with centers P and
B
M
A
Q touch each other at point T
externally. seg BD is a diameter of
Q
the circle with centre Q. line BA is
T
a common tangent touching the
P
other circle at A. Prove that the
D
points D, T and A are collinear. (5 marks)
N
Construction : Draw line MN passing through
point T such that A - M - B and M - T - N
and draw seg BT.
Proof :
In MAT,
AM = MT
[The lengths of the two tangent
segments to a circle drawn from
an external point are equal]

MAT  MTA
[Isosceles triangle theorem]
82
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Let,
m MAT =
Similarly,
m MBT =
m ATB =
 m ATB =









m MTA = x
......(i)
m MTB = y
......(ii)
m MTA + m MTB
[Angle addition property]
x+y
......(iii) [From (i) and (ii)]
In ATB,
m TAB + m ABT + m ATB = 180º [Sum of the measures of angles of
a triangle is 180º]
x + y + x + y = 180 [From (i), (ii) and (iii)]
2x + 2y = 180
2 (x + y) = 180
x + y = 90
m ATB = 90º ......(iv)
[From (iii)]
m DTB = 90º ........(v)
[Angle subtended by a semicircle]
Adding (iv) and (v),
m ATB + m DTB = 90 + 90
m ATB + m DTB = 180º
Also ATB and DTB are adjacent angles.
ATB and DTB form a linear pair [Converse of linear pair axiom]
ray TA and ray TD are opposite rays
Points D, T and A are collinear.
PROBLEM SET - 2 (TEXT BOOK PAGE NO.193)
5.
A
In a right angled triangle ABC, ACB = 90º a
circle is inscribed in the triangle with radius r.
a, b, c are the lengths of the sides BC, AC and
AB respectively. Prove that 2r = a + b – c.
(4 marks)
Q
P
O
C
B
R
Let the centre of the inscribed circle be ‘O’
Let AP = AQ = x
........(i)
[The lengths of the two tangent
CP = CR = y
.......(ii)
segments to a circle drawn from
BR = BQ=
z
......(iii)
an external point are equal]
a + b – c = BC + AC – AB
 a + b – c = CR + RB + AP + PC – (AQ + QB)
[B - R - C, A - P - C , A - Q - B]
 a + b – c = y + z + x + y – (x + z) [From (i), (ii) and (iii)]
 a+b–c = y+z+x+y–x–z
 a + b – c = 2y
 a + b – c = 2y
 a + b – c = 2CP
........(iv) [From (ii)]
In  PCRO
m OPC = m ORC = 90º
[Radius is perpendicular to tangent]
m PCR = 90º
[Given]
 m POR = 90º
[Remaining angle]
 PCRO is a rectangle
[By definition]

CP = OR
........(v) [Opposite sides of a rectangle]
 a + b – c = 2 OR
[From (iv) and (v)]
 a + b – c = 2r
Proof :
S C H O O L S E C TI O N
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PROBLEM SET - 2 (TEXT BOOK PAGE NO.193)
6.
Q
In the adjoining figure, points P, B and Q
are points of contact of the respective
tangents. line QA is parallel to line PC.
If QA = 7.2 cm, PC = 5 cm, find the
radius of the circle.
(5 marks)
A
O
B
C
P
Construction : Draw seg OA and seg OC.
[The
lengths
of
the
two
tangent
Sol.
AQ = AB = 7.2 cm
.....(i)
CP = CB = 5 cm
....(ii) segments to a circle drawn from
an external point are equal]
In OQA and OBA,
m OQA = mOBA = 90º
[Radius is perpendicular to tangent]
Hypotenuse OA  Hypotenuse OA
[Common side]
seg OQ  seg OB
[Radii of same circle]
 OQA  OBA
[By Hypotenuse side theorem]
 AOQ  AOB
[c.a.c.t.]
Let,
m AOQ = m AOB = xº
.....(iii)
[Angle addition property]
m QOB = m AOQ + m AOB
m QOB = x + x
[From (iii)]
 m QOB = 2xº
......(iv)
Similarly,
m COP = m COB = yº
......(v)
m POB = 2yº
.....(vi)
m AOC = m AOB + m COB
[Angle addition property]
m AOC = x + y
....(vii) [From (iii) and (v)]
m QOB + m POB = 180º
[Linear pair axiom]

2x + 2y = 180

2 (x + y) = 180

x + y = 90º

m AOC = 90º
.....(viii)[From (vii)]
In AOC,
m AOC = 90º
[From (viii)]
seg OB  Hypotenuse AC
[Radius is perpendicular to tangent]
 OB2 = AB × BC
[By property of geometric mean]
 OB2 = 7.2 × 5
[From (i) and (ii)]
 OB2 = 36
 OB = 6cm
[Taking square roots]
 Radius of the circle is 6 cm.
EXERCISE - 2.1 (TEXT BOOK PAGE NO.52)
In the adjoining figure,
BC and BA are tangents to circle.
Prove that OD is perpendicular
bisector of AC, where O is the
centre of the circle.
(3 marks)
Proof :
C
11.
OA = OC
BC = BA
84
O
D
B
A
........(i)
[Radii of same circle]
.......(ii)
[The lengths of the two tangent
segments to a circle drawn from an external point are equal]
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 Points O and B are equidistant from the end points A and C of seg AC.
[From (i) and (ii)]
 Points O and B lie on the perpendicular bisector of seg AC.
[By perpendicular bisector theorem]
 seg OB is the perpendicular bisector of seg AC.
 seg OD is the perpendicular bisector of seg AC.
[ O - D - B]
THEOREM
If two circles are touching circles then the common point lies on the
line joining their centres.
Externally touching circles :
In the adjoining figure,
two circles with centres O
and A are touching externally at point P.
O-P-A
Internally touching circles :
In the adjoining figure,
two circles with centres O
and A are touching internally at point P.
O-A-P
P
O
O
A
A
P
EXERCISE - 2.2 (TEXT BOOK PAGE NO.56)
1.
If two circles touch externally then show that the distance between
their centers is equal to the sum of their radii.
(2 marks)
T
Given :
Two circles with centres O and A
O
touch each other externally at point T.
To Prove : OA = OT + AT
Proof :
O-T-A
[If two circles are touching
circles then the common point
lies on the line joining their centres]
 OA = OT + AT
[ O - T - A]
A
EXERCISE - 2.2 (TEXT BOOK PAGE NO.56)
2.
Sol.
If two circles with radii 8 and 3 respectively touch internally then
show that the distance between their centers is equal to the difference
of their radii, find that distance.
(2 marks)
Let two circles with centres O and
A touch each other internally at point T.
 O-A-T
[If two circles are touching
circles then the common
O
T
point lies a the line joining
A
their centres]
 OT = OA + AT
[O - A - T]
 OA = OT – AT
OT = 8 units, AT = 3 units
[Given]
 OA = 8 – 3
 OA = 5 units

 The distance between the centres is 5 units.
S C H O O L S E C TI O N
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EXERCISE - 2.2 (TEXT BOOK PAGE NO.56)
5.
Sol.
Three congruent circles with
centres A, B and C and with
A
radius 5 cm each, touch each
other in points D, E, G as shown in
(i) What is the perimeter of ABC ?
F
(ii) What is the length of side DE of DEF ? (3 marks)
A - D - B [If two circles are touching circles
B - E - C then the common point lies on the
A - F - C line joining their centres]




D
B
E
C
AD = DB = BE = EC = AF = FC = 5cm .......(i)
[Radii of congruent circles and given]
AB = AD + BD
[ A - D - B]
AB = 5 + 5
[From (i)]
AB = 10 cm
......(ii)
Similarly, BC = 10 cm ......(iii)
AC= 10 cm
.......(iv)
Perimeter of ABC = AB + BC + AC
= 10 + 10 + 10 [From (ii), (iii) and (iv)]
Perimeter of ABC = 30 cm
(ii) In ABC,
D and E are mid-points of sides AB and BC respectively.
[From (i)]
1
AC
[By mid-point theorem]
DE =
2
1
 DE =
× 10
[From (iv)]
2

DE = 5 cm.
EXERCISE - 2.2 (TEXT BOOK PAGE NO.56)
3.
Two circles with centers A, B are touching externally and a circle with centre
C touches both externally. Suppose AB = 3 cm, BC = 3 cm, CA = 4 cm.
Find the radii of all circles.
(4 marks)
Given : Circles with centres A, B and C
touch each other pairwise externally
B
A
P
at points P, Q and R respectively.
Q
AB = 3 cm, BC = 3cm, CA = 4cm.
R
To Find : Radii of the circles with centre A, B and C.
Sol.
A-P-B
[If two circles are touching circles
C
B-Q-C
then the common point lies on the
line joining their centres]
A-R-C
Let,AP = AR = x
BP = BQ = y
[Radii of the same circle]
CQ = CR = z
AP + BP
= AB
[ A - P - B]
 x+y
= 3
........(i)
BQ + CQ = BC
[B - Q - C]
 y+z
= 3
.......(ii)
AR + CR = AC
[ A - R - C]
 x+z
= 4
......(iii)
86
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

Adding (i), (ii) and (iii),
x+y+y+z+x+z = 3+3+4
2x + 2y + 2z = 10
2(x + y + z) = 10
x + y + z = 5 ......(iv)
Substituting (i) in (iv),
3+z = 5
z = 5–3
z = 2
Substituting (ii) in (iv),
x+3= 5
x = 5–3
x = 2
Substituting (iii) in (iv),
y+4= 5
y = 5–4
y = 1

Radii of circles with centres A, B and C are 2 cm, 1 cm and 2 cm respectively.







EXERCISE - 2.2 (TEXT BOOK PAGE NO.57)
8.
In the adjoining figure,
A and B are centers of two
circles touching each other at M.
Line AC and line BD are tangents.
If AD = 6 cm and BC = 9 cm then find
the length of seg AC and seg BD. (3 marks)
Sol.
A-M-B
C
D
A
M
B
[If two circles are touching circles then
the common point lies on the line
joining their centres]
........(i)
[Radii of the same circle]
........(ii)
[ A - M - B]
[From (i) and (ii)]
........(iii)






AM = AD = 6 cm
BM = BC = 9 cm
AB = AM + MB
AB = 6 + 9
AB = 15 cm
In ABC,
m ACB = 90º
AB² = AC² + BC²
15² = AC² + 9²
225 = AC² + 81
AC² = 225 – 81
AC² = 144
AC = 12 cm
In ADB,
m ADB = 90º
AB 2 = AD2 + BD2
152 = 62 + BD2
225 = 36 + BD2
BD2 = 225 – 36
BD2 = 189
BD =
9  21

BD

The lengths of seg AC and seg BD are 12 cm and 3 21 cm respectively.








=
S C H O O L S E C TI O N
3 21 cm.
[Radius is perpendicular to the tangent]
[By Pythagoras theorem]
[From (ii) and (iii)]
[Taking square roots]
[Radius is perpendicular to the tangent]
[By Pythagoras theorem]
[From (i) and (iii)]
[Taking square roots]
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EXERCISE - 2.2 (TEXT BOOK PAGE NO.56)
4.
D
In the adjoining figure,
points P and Q are the centers of
C
the circles. Radius QN = 3, PQ = 9.
M is the point of contact of the circles.
N
M
Q
P
Line ND is tangent to the larger circle.
Point C lies on the smaller circle.
Determine NC, ND and CD.
(5 marks)
Construction : Draw seg CM and seg DP
Sol.
Q-M-P
[If two circles are touching circles then
the common point lies on the line joining their centre]
QN = 3 units
[Given]
PQ = 9 units
PN = PQ + QN
[ P - Q - N]

PN = 9 + 3

PN = 12 units
QN = QM = 3 units
[Radii of the same circle]

MN = 6 units
[ Diameter is twice the radius]
PM + QM = PQ
[ P - M - Q]

PM + 3 = 9
[Given]

PM = 9 – 3

PM = 6 units
PM = PD = 6 units
[Radii of the same circle]
In NDP,
m NDP = 90º
......(i) [Radius is perpendiculer to the tangent]

NP² = ND² + PD²
[By Pythagoras theorem]

12² = ND² + 6²

144 = ND² + 36

ND² = 144 – 36

ND² = 108

ND =

NC = CD =
108

ND =
36  3

ND = 6 3 units
Now, m NCM = 90º
.....(ii) [Angle inscribed in a semicircle
is a right angle]
NCM  NDP
[From (i) and (ii)]
 seg CM || seg DP
....(iii) [By corresponding angles test]
In NDP,
seg CM || seg DP,
[From (iii)]
NC
NM

=
[By B.P.T.]
CD
MP
NC
6

=
CD
6
NC

= 1
CD

NC = CD
 C is the midpoint of seg ND
1

NC = CD = ND
2
1
6 3

NC = CD =
2
88
3 3 units.
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PROBLEM SET - 2 (TEXT BOOK PAGE NO.193)
4.
In the adjoining figure, two circles touch
D
P
each other internally in a point A. The radius
of the smaller circle with centre M is 5.
A
C
The smaller circle passes through the centre N
N
M
of the larger circle. The tangent to the smaller
circle drawn through C intersects the larger circle
(4 marks)
in point D. Find CD.
Construction : Draw seg AD.
Sol.
NM = MA = MP = 5 units
[Radii of same circle]
NA = 2 NM
[Diameter is twice the radius]

NA = 2 × 5

NA = 10 units
CN = NA = 10 units
[Radii of same circle]
CA = 2 CN
[Diameter is twice the radius]
CA = 2 × 10
CA = 20 units
CM = CN+ NM
[C - N - M]

CM = 10 + 5

CM = 15 units
m CPM = 90º
[Radius is perpendicular to tangent]
m CDA = 90º
[Angle subtended by a semicircle]
 CPM  CDA
 seg PM || seg DA
......(i) [By corresponding angles test]
In CPM
CPM = 90º
[Radius is perpendicular to the tangent]
2
2
2
CM = CP + PM
[By Pythagoras theorem]
152 = CP2 + 52
225 = CP2 + 25
 CP 2 = 225 – 25
 CP 2 = 200

CP =
[Taking square roots]
100 × 2

CP = 10 2 units
In CDA,
seg PM || seg DA
[From (i)]
CP
CM

=
[By B.P.T.]
PD
MA
15
10 2

=
5
PD





PD =
10 2 × 5
15
10 2
units
3
CD = CP + PD
PD =
[C - P - D]
10 2 10 2

1
3
30 2  10 2
CD =
3
CD =
CD =
S C H O O L S E C TI O N
40 2
units
3
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EXERCISE - 2.2 (TEXT BOOK PAGE NO. 56)
7.
Explain what is wrong in the
information provided in the
adjacent figure (do not take
actual measurements).
OP = 4, PB = 4, BQ = 5.
(2 marks)
Sol.






P
4
B
4
O
Q
Let the two circles touch at point M.
O-M-Q
[If two circles are touching circles
then the common point lies on the joining their centres]
OM = OP = 4 units
[Radii of same circle]
MQ = BQ = 5 units
OQ = OM + MQ
[O - M - Q]
OQ = 4 + 5
OQ = 9 units
In OBQ
m OBQ = 90º
[Radius is perpendicular to tangent]
OQ 2 = OB2 + BQ2
[By Pythagoras theorem]
OQ 2 = 82 + 52
OQ 2 = 64 + 25
OQ 2 = 89
 OQ = 89 units
[Taking square roots]
But one segment cannot have two different lengths.
EXERCISE - 2.2 (TEXT BOOK PAGE NO. 56)
6.
Sol.
Two circles which are not congruent touch externally. The sum of their
areas is 130cm2 and the distance between their centers is 14 cm. Find
radii of circles.
(3 marks)
Let the radius of first circle be r1 and that of second circle be r2
Circles are touching externally

r1 + r2 = 14 cm

r 2 = 14 – r1 ........(i)
According to given information
A (I Circle) + A (II Circle) = 130  cm2

r12 + r22 = 130 

 (r12 + r22) = 130 
r12 + r22 = 130

2

r1 + (14 – r1)2 = 130
[From (i)]
2

r1 + 196 – 28r1 + r12 = 130
 2r12 – 28r1 + 196 – 130 = 0

2r12 – 28r1 + 66 = 0

2 (r12 – 14r1 + 33) = 0

r12 – 14r1 + 33 = 0
2

r1 – 11r1 – 3r1 + 33 = 0
 r1 (r1 – 11) – 3 (r1 – 11) = 0
(r1 – 3) (r1 – 11)
= 0
r1 – 3 = 0
or
r1 – 11 = 0

r 1 = 3 units
or
r 1 = 11 units
If r1 = 3
If r1 = 11
then,r 2 = 14 – 3
then, r2 = 14 – 11
r 2 = 11 units
r 2 = 3 units
 The radii of the circles is 11 units and 3 units.
90
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PART - II : CIRCLE - ARC
•
•
An arc is a part of a circle.
X
•
If a secant passes through the centre,
then each arc so formed is called a semicircle. A
In the adjoining figure, arc AXB and arc AYB
are semicircles.
B
•
O
•
•
An arc smaller than the semicircle
is called a minor arc.
•
An arc greater than the semicircle is called a major arc. A•
In the adjoining figure, arc AXB is a minor arc
and AYB is the corresponding major arc.
•
Y
An angle in the plane of the circle with its vertex
at the centre is called a central angle.
•X
•B
•
O
•
Y
In the adjoining figure,
AOB is the central angle corresponding
to minor arc AXB.
O
(
•
A
•
B
•
Measure of a minor arc is equal to the measure
of its corresponding central angle.
In the adjoining figure, m (arc AXB) = m AOB
•
Measure of major arc = 3600 – measure of corresponding minor arc.
•Y
In the adjoining figure,
m(arc AYB) = 3600 – m(arc AXB)
•
•
`
Measure of a circle is 3600
Measure of a semicircle is 1800
O
•
A•
•
•B
X
Arc addition property :
In the adjoining figure,
arc APB and arc BQC have in
common only the end point B,
\ m(arc APB) + m(arc BQC) = m(arc ABC)
`
X
B
•
P
•
O
•
A•
•Q
•C
Congruent arcs :
Two arcs of the same circle or of congruent circles, having equal measures
are congruent.
`
Inscribed angle :
An angle is said to be an inscribed angle, if
B
(i) the vertex is on the circle and
(ii) both the arms are secants.
In the adjoining figure,
•
ABC is an inscribed angle,
because vertex B lies on the
A
circle and both the arms BA
and BC arc secants.
In other words, ABC is inscribed in arc ABC.
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Intercepted arc :
Given an arc of the circle and an angle, if each side of the angle contains an
end point of the arc and all other points of the arc except the end points lie
in the interior of the angle, then the arc is said to be intercepted by the angle.
A
A
D
A
B
B
D
C
B
C
E
C
(a)
(b)
(c)
B
A
B
B
A
C
C
C
A
(d)
(e)
(f)
In figures (a), (b) and (c), ABC has its vertex B outside the circle and
intercepts two arcs.
(ii) In figures (d) and (e), ABC has its vertex on the circle and intercepts
only one arc.
(iii) In figure (f), ABC has its vertex B inside the circle and intercepts only
one arc.
(i)
INSCRIBED ANGLE THEOREM
The measure of an inscribed angle is half of the measure of its
B
intercepted arc.
In the adjoining figure,
ABC is an inscribed angle and
arc AXC is the intercepted arc.
1
m ABC =
m (arc AXC)
2
•
X
Corollary - 1 :


92
A
An angle inscribed in a semicircle is a right angle.
In the adjoining figure,
A
ABC is inscribed in
the semicircle ABC.
m ABC = 90º
C
B
•
O
C
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Corollary - 2 :

B
D
Angles inscribed in the same arc are congruent.

In the adjoining figure,
ABC and ADC, both are
inscribed in the same arc ABC.
ABC  ADC.
A
C
•
X
 NOTE 

Inscribed angles, ABC and ADC both intercept the same arc AC.
ABC  ADC.
EXERCISE - 2.3 (TEXT BOOK PAGE NO. 62)
1.
Q
•
B
In the adjoining figure,
if m (arc APC) = 60º
and m BAC = 80º
Find (a) ABC (b) m (arc BQC). (2 marks)
Sol.
C
80º
(a)


1
m ABC =
m(arc APC)
2
1
m ABC =
× 60
2
•P
[Inscribed angle theorem]
m ABC = 30º
1
m(arc BQC)
2
1

80 =
m(arc BQC)
2
 m(arc BQC) = 80 × 2
(b)
A
m BAC =
[Inscribed angle theorem]
 m(arc BQC) = 160º
EXERCISE - 2.3 (TEXT BOOK PAGE NO. 62)
3.
Chords AB and CD of a circle
intersect in point Q in the interior
of a circle as shown in figure,
if m (arc AD) = 25º circle and
m (arc BC) = 31º, then find BQC. (2 marks)
Construction : Draw seg AC.
Sol.
m (arc AD) = 25º
[Given]
m ACD =
1
m(arc AD)
2
1
× 25º
2

m ACD = 12.5º

m ACQ = 12.5º
m (arc BC) = 31º

D
A
B
Q
C
[Inscribed angle theorem]
m ACD =
S C H O O L S E C TI O N
[ D - Q - C]
[Given]
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





1
m(arc BC)
2
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[Inscribed angle theorem]
1
× 31º
2
m BAC = 15.5º
m QAC = 15.5º
[ A - Q - B]
BQC is an exterior angle of AQC,
m BQC = m QAC + m ACQ [Remote interior angle theorem]
m BQC = 15.5º + 12.5º
m BAC =
m BQC = 28º
EXERCISE - 2.3 (TEXT BOOK PAGE NO. 62)
4.
Sol.
Secants containing chords RS
and PQ of a circle intersects each
other in point A in the exterior of
a circle, as shown in figure If
A
m (arc PCR) = 26º and
m (arc QDS) = 48º
then find
(i) AQR (ii) SPQ (iii) RAQ (3 marks)
(i) m (arc PR) = 26º
[Given]
S
R
•D
C
P
Q
m AQR =
1
m (arc PCR)
2
[Inscribed angle theorem]

m AQR =
1
× 26º
2
[Given]

m AQR = 13º
(ii) m (arc QS) = 48º
[Given]
m SPQ =
1
m (arc QDS)
2
[Inscribed angle theorem]

m SPQ =
1
× 48º
2
[Given]

m SPQ = 24º
(iii)
m SRQ =
1
m (arc QDS)
2
[Inscribed angle theorem]

m SRQ =
1
× 48º
2
[Given]

m SRQ = 24º
SRQ is an exterior angle of ARQ,

m SRQ = m RAQ + m AQR [Remote interior angle theorem]

24º = m RAQ + 13º

m RAQ = 24º – 13º

94
m RAQ = 11º
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EXERCISE - 2.3 (TEXT BOOK PAGE NO. 62)
5.
In the adjoining figure,
D
C
in two chords AB and CD
A
B
of the same circle are parallel
to each other. P is the centre
P
of the circle.
Prove : m CPA = m DPB. (3 marks)
Construction : Draw seg BC.
Proof : m CPA = m (arc CA)
.......(i)
[Definition of measure of minor arc]
......(ii)
m DPB = m (arc DB)
m ABC =
1
m (arc CA)
2
......(iii)
1
m (arc DB) ......(iv)
2
chord CD || chord AB
On transversal BC,
ABC  BCD
......(v)
1
1
m (arc CA) =
m(arc DB)
2
2
m (arc CA) = m (arc DB)
......(vi)
m CPA = m DPB
[Inscribed angle theorem]
m BCD =




[Given]
[Converse of alternate angles test]
[From (iii), (iv) and (v)]
[From (i), (ii) and (vi)]
EXERCISE - 2.3 (TEXT BOOK PAGE NO. 62)
6.
In the adjoining figure,
m (arc XAZ) = m (arx YBW).
Prove that XY || ZW
(2 marks)
X
Y
A
Construction : Draw seg XW
Z
1
Proof :
m XWZ =
m (arc XAZ) .......(i)
2
1
m (arc YBW) .......(ii)
m WXY =
2
But, m (arc XAZ) = m (arc YBW)
 m XWZ = m WXY
 line XY || line ZW
B
W
[Inscribed angle theorem]
.......(iii)
[Given]
[From (i), (ii) and (iii)]
[Alternate angles test]
EXERCISE - 2.3 (TEXT BOOK PAGE NO. 62)
7.
Given circle with centre O
and BC || ED, m (arc BC) = 94º,
m (arc ED) = 86º, ADE = 8º
Find (i) m (arc AE), (ii) m (arc DC),
(iii) m (arc EB).
Also find DAB, ECB, CBE. (5 marks)
Construction : Draw seg BD
Proof :
m BDE =
1
m (arc BAE)
2
1
m CBD =
m (arc CD)
2
line BC || line ED
S C H O O L S E C TI O N
B
C
E
D
A
.....(i)
[Inscribed angle theorem]
.....(ii)
[Given]
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GEOMETRY
On transversal BD
CBD  BDE
......(iii)


1
1
m (arc BAE) =
m (arc CD)
2
2
m (arc BAE) = m (arc CD)
m ADE =
1
m (arc AE)
2
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[Converse of alternate angles test]
[From (i), (ii) and (iii)]
.......(iv)
[Inscribed angle theorem]
1
m (arc AE)
2

m (arc AE) = 16º
m (arc AD) = m (arc AE) + m (arc ED)
[Arc Addition property]

86º = m (arc ED) + 16º

m (arc ED) = 86 – 16

m (arc ED) = 70º
.......(iii)
m (arc BC) + m (arc CD) + m (arc BAE) + m (arc ED) = 360º
[Measure of a circle is 360º]

94º + m (arc BAE) + m (arc BAE) + 70º = 360º
[From (ii), (iii) and given]

2 m (arc BAE) + 164 = 360

2 m (arc BAE) = 360 – 164

2 m (arc BAE) = 196
8º =

m (arc BAE) = 98º

m (arc BCD) =
=
 m (arc BCD) =
m (arc DC) = 98º
[From (iv) and (v)]
m (arc BC) + m (arc CD)
[Arc Addition property]
94 + 98
192º
1
m (arc BCD)
[Inscribed angle theorem]
2
m DAB
=

m DAB
=

m DAB
= 96º
m ECB
=
1
m (arc BAE)
2
=
1
× 98º
2
m  ECB
1
× 192º
2
[Inscribed angle theorem]
= 49º
m (arc CDE) = m (arc CD) + m (arc DE)
= 98 + 70
m (arc CDE) = 168º
96
........ (v)
m CBE
=
1
m (arc CDE)
2

m CBE
=
1
× 168º
2

m CBE
= 84º
[Arc addition property]
[Inscribed angle theorem]
S C H O O L S E C TI O N
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EXERCISE - 2.3 (TEXT BOOK PAGE NO. 62)
2.
Find the radius of the circle passing through the vertices of a right angled
(3 marks)
triangle when lengths of perpendicular sides are 6 and 8.
Given : (i) In PQR, m PQR = 90º
P
(ii) PQ = 6 units, QR = 8 units.
6
Q
(iii) Points P, Q and R lie on the circle.
To Find : radius of the circle.
Sol.
In PQR,
•Y
8
[Given]
m PQR = 90º
 PR² = PQ² + QR²
[By Pythagoras theorem]
R
 PR² = 6² + 8²
 PR² = 36 + 64
 PR² = 100
 PR = 10 units
[Taking square roots]
Let Y be a point on the circle as shown in the figure
[Given]
m PQR = 90º
1
m(arc PYR)
[Inscribed angle theorem]
m PQR =
2
1

90º =
m(arc PYR)
2
 m (arc PYR) = 180º
 arc PYR is a semicircle
 seg PR is the diameter.
 Diameter = 10 units.

Radius = 5 units
[ Radius is half of the diameter]


Radius of the circle is 5 units.
EXERCISE - 2.5 (TEXT BOOK PAGE NO. 75)
1.
Let two circles intersect each at points A and D. Let the diameter AB
intersect the circle with centre P at point N and diameter AC intersects
the circle in point M with centre Q. Then prove that AC. AM = AB. AN.
(3 marks)
A
N
Q
P
M
Construction : Draw seg BM and seg CN
B
DC
Proof :
seg AB is the diameter
 m AMB = 90º
[Angle subtended by a semicircle]
seg AC is the diameter
 m ANC = 90º
[Angle subtended by a semicircle]
 m AMB = mANC
.......(i)
In AMB and ANC
AMB  ANC
[From (i)]
BAM  CAN
[Common angle]
 AMB ~ ANC
[By AA test of similarity]
AM
AB
=
AN
AC
 AC.AM = AB.AN

S C H O O L S E C TI O N
[c.s.s.t]
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PROBLEM SET - 2 (TEXT BOOK PAGE NO. 194)
A
ABC is inscribed in a circle with centre O, seg AX
is a diameter of the circle with radius r.
seg AD  seg BC. Prove that
abc
O
(i) ABX ~ ADC, (ii) A (ABC) =
.
4r
B
D
C
(4 marks)
(a is side opposite to A, ...)
Proof :
seg AX is a diameter
[Given]
X
 arc ACX is a semi circle.
 m ABX = 90º
[Angle subtended by a semicircle]
In ABX and ADC
ABX  ADC
[Each is 90º]
AXB  ACD
[Angles inscribed in the same arc
and C - D - B]
 ABX ~ ADC
[By AA test of similarity]
BC = a, AB = c, AC = b
[Given]
AB
AX
=
[c.s.s.t.]
AD
AC
c
2r
=
AD
b
bc

AD =
.......(i)
2r
1
A (ABC) =
× base × height
2
1
× BC × AD
=
2
1
bc
=
×a×
2
2r
11.
 A (ABC) =
abc
4r
 Cyclic Quadrilateral :
A
D
A quadrilateral whose all the four vertices lie on a
circle is called cyclic quadrilateral.
In the adjoining figure,
ABCD is cyclic,
as all the four vertices A, B, C and D.
lie on a circle.
B
C
THEOREM
Statement : The opposite angles of a cyclic quadrilateral are supplementary.
Given : ABCD is a cyclic
A
To Prove : m ABC + m ADC = 180º
m BAD + m BCD = 180º
Proof: m ABC =
1
m (arc ADC) .....(i)
2
1
m (arc ABC) .....(ii)
2
Adding (i) and (ii), we get
m ADC =
98
B
[Inscribed angle
theorem]
D
C
S C H O O L S E C TI O N
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1
1
m (arc ADC) +
m (arc ABC)
2
2
1
m ABC + m ADC =
[m (arc ADC) + m (arc ABC)]
2
1
m ABC + m ADC =
× 360º
[ Measure of a circle is 360º]
2
m ABC + m ADC = 180º
..........(iii)
In ABCD,
m BAD + m BCD + m ABC + m ADC = 360º [ Sum of measure of
angles of a quadrilateral is 360º]
m BAD + m BCD + 180º = 360º
[From (iii)]
m BAD + m BCD = 180º
m ABC + m ADC =





THEOREM : If opposite angles of a quadrilateral are supplementary, then
the quadrilateral is cyclic.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 194)
D
If two consecutive angles of cyclic quadrilateral are A
congruent, then prove that one pair of opposite sides
is congruent and other is parallel. More precisely.
Given : ABCD is cyclic quadrilateral in which
ABC BCD. To prove side DC  side AB, AD || BC. B
C
(3 marks)
Given : ABCD is cyclic and ABC  BCD
To Prove : (i) seg AD || seg BC (ii) side DC  side AB.
Construction : Draw seg AC and seg BD
Proof :
ABCD is cyclic
[Given]
 m ABC + m ADC =180º
.......(i) [Opposite angles of a cyclic
quadrilateral are supplementary]
But, ABC  BCD
.......(ii) [Given]
 m BCD + mADC = 180º
[From (i) and (ii)]
 seg AD || seg BC
[Interior angles test]
In ABC and DCB
seg BC  seg BC
[Common side]
ABC  DCB
[Given]
BAC  CDB
[Angles inscribed in same arc]

ABC  DCB
[By SAA test]

side AB  side DC
[c.s.c.t.]
7.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 194)
19.
Prove that in a cyclic trapezium angles at the base are congruent.
(3 marks)
A
D
Given :
ABCD is a cyclic trapezium.
seg AB || seg BC
To prove : ABC  DCB
Proof :
ABCD is cyclic
 m ABC + m ADC = 180º
.....(i)

seg AD || seg BC
On transversal DC,
m DCB + m ADC = 180º
S C H O O L S E C TI O N
B
C
[Given]
[Opposite angles of a cyclic
quadrilateral are supplementary]
[Given]
......(ii) [Converse of interior angle test]
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


m ABC + m ADC = m DCB + m ADC
m ABC = m DCB
ABC  DCB
[From (i) and (ii)]
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 195)
In a cyclic quadrilateral ABCD,
the bisectors of opposite angles A
and C meet the circle at P and Q
respectively. Prove that PQ is a
diameter of the circle. (4 marks)
Proof :
DAP  BAP
Let m DAP = m BAP = xº
....(i)
DCQ  BCQ
Let, m DCQ = m BCQ = yº .....(ii)
ABCD is cyclic
 m DAB + m DCB = 180º
18.










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A
P
D
× C
×
•
•
B
Q
[ ray AP bisects DAB]
[ ray CQ bisects DCB]
[Given]
[Opposite angles of a cyclic
quadrilateral are supplementary]
m DAP + m BAP + DCQ + m BCQ = 180º
[Angle addition property]
x + x + y + y = 180º
[From (i) and (ii)]
2x + 2y = 180º
.....(iii)
1
m DAP =
m (arc DP)
[Inscribed angle theorem]
2
1
x =
m (arc DP)
[From (i)]
2
m (arc DP) = 2xº
.....(iv)
1
m (arc DQ)
[Inscribed angle theorem]
m DCQ =
2
1
y =
m (arc DQ)
[From (ii)]
2
m (arc DQ) = 2yº
......(v)
m (arc DP) + m (arc DQ) = 2x + 2y
[Adding (iv) and (v)]
m (arc PDQ) = 180º
[Arc addition property and from (iii)]
Arc PDQ is a semicircle
seg PQ is a diameter of the circle.
 Converse of cyclic quadrilateral theorem :
If the opposite angles of a quadrilateral are supplementary
D
then it is a cyclic quadrilateral.
A
In ABCD if m A + m C = 180º
or
m B + m D = 180º
then, ABCD is cyclic quadrilateral.
 Corollary :
B
C
An exterior angle of cyclic quadrilateral is congruent to the angle opposite
D
A
to adjacent interior angle.
In the adjoining figure,
ABCD is cyclic
DCE is an exterior angle.
•
B
C E
 DCE  BAD
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 NOTE 
Angle opposite to adjacent interior angle is also called interior opposite angle.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 195)
13.
Prove that the quadrilateral formed by the
D
angle bisectors of a cyclic quadrilateral A
(5 marks)
is also cyclic.
••
Q
R
Given : (i) ABCD is cyclic.
P
S
(ii) Ray AP, ray BQ, ray CR and
C
×
ray DS are the bisectors of
×
B
A, B, C and D respectively.
To Prove : PQRS is cyclic.
Proof :
DAP BAP
[ ray AP bisects BAD]
Let, m DAP = m BAP = aº ........(i)
Similarly,
m ABP = m CBP = bº
........(ii)
.......(iii)
m BCR = m DCR = cº
........(iv)
m ADR = m CDR = dº
In BQC,
m BQC + m QBC + m QCB = 180º [ Sum of the measures of
angles of a triangle is 180º]

m BQC + b + c = 180
[From (ii) and (iii)]

m BQC = (180 – b – c)º

m PQR = (180 – b – c)º
.......(v)
[B - P - Q and C - R - Q]
Similarly, we can prove
......(vi)
m PSR = (180 - a - d)º
Adding (v) and (vi),
m PQR + m PSR = 180 – b – c + 180 – a – d

m PQR + m PSR = 360 – a – b – c – d

m PQR + m PSR = 360 – (a + b + c + d)
.......(vii)
In ABCD,
m BAD + m ABC + m BCD + m ADC = 360º
[ Sum of the measures of angles of a quadrilateral is 360º]
m BAP + m DAP + m ABP + m PBC +
= 360
m BCQ + m DCQ + m CDR + m ADR
[Angle addition property]

a + a + b + b + c + c + d + d = 360
[From (i), (ii), (iii) and (iv)]

2a + 2b + 2c + 2d = 360

2 (a + b + c + d) = 360

a + b + c + d = 180
......(viii)

m PQR + m PSR = 360 – 180
[From (vii) and (viii)]

m PQR + m PSR = 180º
 PQRS is cyclic.
[If opposite angles of a quadrilateral
are supplementary, then quadrilateral is cyclic]
S C H O O L S E C TI O N
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PROBLEM SET - 2 (TEXT BOOK PAGE NO. 195)
15.
If all the sides of a quadrilateral are produced in the same order (clockwise
or anticlockwise). Show that quadrilateral formed by the bisectors of
exterior angles is cyclic.
(5 marks)
L
Given : (i) ABCD is cyclic
Q
(ii) Lines PQ, QR, RS and PS are
the bisectors of the LAB, MBC, NCD,
×
B
A ×
KDA respectively.
•
M
•
To prove : PQRS is cyclic
P
Proof :
Let m LAQ = m BAQ = aº .......(i)
R
......(ii)
m MBR = m CBR = bº
K
(
m NCS = m DCS = cº
.....(iii)
D
 C
......(iv)
m KDP = m ADP = dº
[Lines PQ, QR RS and PS are
S
the bisectors]
N
m SPQ = 180 – a – d
Similarly,
m QRS = 180 – b – c
......(vi)
.....(vii)
m PQR = 180 – a – b
m PSR = 180 – c – d
....(viii)
Adding (v) and (vi),
m SPQ + m QRS = 180 – a – d + 180 – b – c
m SPQ + m QRS = 360 – (a + b + c + d)
......(ix)
In PQRS,
m SPQ + m PQR + m QRS + m PSR = 360º
[Sum of the measures of angles of quadrilateral is 360º]
 180 – a – d + 180 – a – b + 180 – b – c + 180 – c – d = 360

720 – 2a – 2b – 2c – 2d = 360

720 – 360 = 2a + 2b + 2c + 2d

360 = 2a + 2b + 2c + 2d

a + b + c + d = 180º
.....(x)
[Dividing throughout by 2]

m SPQ + m QRS = 360º – 180º
[From (ix) and (x)]

m SPQ + m QRS = 180º
 PQRS is cyclic
[If the opposite angles of quadrilateral
are supplementary then it is cyclic]

(
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 195) A
In ABC, midpoints of sides AB,
AC and BC are P, Q and R respectively.
AS  BC. Prove that PQRS is a
P
(5 marks)
cyclic quadrilateral.
Proof :
In ABC,
B
S
P and Q are midpoint of seg AB and seg AC
 seg PQ || seg BC
[By midpoint theorem]
 seg PQ || seg BR
[B - R - C]
21.
102
Q
R
C
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Similarly, seg QR || seg PB
 PBRQ is a parallelogram
In ASB,
m ASB = 90º
seg SP is median to hypotenuse AB

SP =
1
AB
2
[By definition]
[Given]
[Given]
.....(i) [In a right angled triangle the median
drawn to the hypotenuse is half of it]
But,
PB =
1
AB
2
.....(ii) [ P is the midpoint of side AB]
In PBS,
SP = PB
m PBS = m PSB
PBRQ is a parallelogram

m PBR = m PQR
[From (i) and (ii)]
[Isosceles triangle theorem]



m PBS = m PQR
m PSB = m PQR
[Opposite angles of a parallelogram
are congruent]
.....(iv) [B - S - R]
......(v) [From (iii) and (iv)]
But,
[Linear pair axiom]
m PSB + m PSR = 180º
m PQR + m PSR = 180º
[From (v)]
 PQRS is cyclic
[If opposite angles of a quadrilateral are
supplementary then it is a cyclic quadrilateral]
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 195)
16.
P
In the adjoining figure,
in the isosceles triangle PQR,
the vertical P = 50º. The circle
passing through Q and R cuts PQ .
in S and PR in T. ST is joined.
(3 marks)
Find PST.
50º
S
T
Sol.
Q
In PQR
R
[Given]
seg PQ  seg PR
 PQR  PRQ
......(i) [Isosceles triangle theorem]
m PQR + m PRQ + m QPR = 180º [Sum of the measures of angles
of a triangle is 180º]

m PRQ + m PRQ + 50 = 180º[From (i) and Given]
2m PRQ = 180º – 50º

2m PRQ = 130º

m PRQ = 65º
......(ii)
SQRT is cyclic
 PST  TRQ
[An exterior angle of cyclic
quadrilateral is congruent to the
angle opposite to adjacent interior
angle]
 PST  PRQ
[P - T - R]
 PST = 65º
S C H O O L S E C TI O N
[From (ii)]
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PROBLEM SET - 2 (TEXT BOOK PAGE NO. 195)
17.
ABCD is a parallelogram.
A circle passing through D, A,
B cuts BC in P.
Prove that DC = DP. (3 marks)
Proof :
C
P
D
B
A
ABPD is cyclic
 DPC  DAB
.....(i)
[By definition]
[An exterior angle of cyclic quadrilateral
is congruent to the angle opposite to
adjacent interior angle]
ABCD is parallelogram
 DCB  DAB
.......(ii) [Opposite angles of a parallelogram
are congruent]
 DPC  DCB
......(iii) [From (i) and (ii)]
In DPC,
DPC  DCP
[From (ii) and C - P - B]
 seg DP  seg DC
[Converse of isosceles triangle theorem]
 DP = DC
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 194)
8.
Two circles intersect each other in points X and Y. Secants through X
and Y intersect one of the circles in A and D and the other in B and C
respectively. A and B are on opposite sides of line DC. Show that line
A
AD is parallel to line BC.
(3 marks)
X

C
B
Y
Proof :
ADYX is cyclic
 XYC  DAX
But, XYC  XBC
 DAX  XBC
 DAB  ABC
 line AD || line BC
D
[By definition]
......(i) [The exterior angle of a cyclic
quadrilateral is equal to its interior
opposite angle]
.....(ii) [Angles inscribed in the same arc]
[From (i) and (ii)]
[A - X - B]
[By Alternate angles test]
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 195)
20.
Two circles intersect each other
in points A and B. Secants through
A and B intersects circles in C, D and
M, N. Prove that CM || DN. (3 marks)
C
A
D
M
N
B
Construction : Draw seg AB.
Proof :
ABMC is cyclic
[By definition]
 m MCA + m MBA = 180º ......(i) [Opposite angles of a cyclic
quadrilateral are supplementary]
ABND is cyclic
[By definition]
104
S C H O O L S E C TI O N
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 MBA  ADN
.....(ii) [The exterior angle of a cyclic
quadrilateral is equal to its interior
opposite angle]
 m MCA + m ADN = 180º
[From (i) and (ii)]
 m MCD + m CDN = 180º
[C - A - D]
 seg CM || seg DN
[By Interior angles test]
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 194)
12.
ABCD is cyclic quadrilateral, lines AB and DC
intersect in the point F and lines AD and BC
intersect in the point E. Show that the
B
circumcircles of BCF and CDE intersect
in a point G on the line EF.
(4 marks)
A
D
C
F
E
G
Construction : Draw seg CG.
Proof :
Let the circumcircle of BCF and CDE intersect at points C and G.
BCGF is cyclic.
[By definition]
 m ABC = m CGF
.....(i) [Exterior angle of a cyclic quadrilateral
is equal to its interior opposite angle]
DCGE is cyclic
[By definition]
 m ADC = m CGE
......(ii) [Exterior angle of a cyclic quadrilateral
is equal to its interior opposite angle]
Adding (i) and (ii),
m ABC+ m ADC = m CGF + m CGE
.......(iii)
ABCD is cyclic
 m ABC + m ADC = 180º ....(iv) [Opposite angle of a cyclic quadrilateral
are supplementary]
 m CGF + m CGE = 180º
[From (iii) and (iv)]
Also, CGF and CGE are adjacent angle
 CGF and CGE form a linear pair
[Converse of linear pair axiom]
 ray GF and ray GE form opposite rays
 Points F, G, E are collinear.
 G lies on line EF
 The circumcircle of BCF and CDE intersect in a point on line EF.
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 194)
10.
In the adjoining figure,
two circles intersects each other
in points A and B. Secants through the
point A intersect the circles in points P, Q
and R, S. Line PR and line SQ intersect in point T.
Show that
P
(i) PTQ and PBQ are supplementary,
(ii) BSTR is a cyclic quadrilateral.
(5 marks)
Construction : Draw seg AB
Proof :(i) PBAR is cyclic
[By definition]

TRA  ABP

TRS  ABP
S C H O O L S E C TI O N
T
R
A
Q
S
B
[The exterior angle of a cyclic quadrilateral
is equal to its interior opposite angle]
......(i) [R - A - S]
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



EDUCARE LTD.
ABQ  ASQ
[Angles inscribed in same arc]
ABQ  RST
.....(ii) [R - A - S and S - Q - T]
In TRS,
m TRS + m RST + m RTS = 180º
[Sum of measures of angles of
a triangle is 180º]
m ABP + m ABQ + m RTS = 180º
[From (i) and (ii)]
m PBQ + m RTS = 180º
[Angle addition property]
m PBQ + m PTQ = 180º [Q P - R - T and T - Q - S]
T
(ii)
R
A
Q
P
S
B

ABR  APR
ABR  QPT
ABSQ is cyclic
TQA  ABS
[Angles inscribed in the same arc]
......(i) [P - R - T and P - A - Q]
[By definition]
[The exterior angle of a cyclic quadrilateral
is equal to its interior opposite angle]
......(ii) [P - A - Q]
TQP  ABS
In TPQ,
m QPT + m TQP + m PTQ = 180º




[Sum of measures of angles of
a triangle is 180º]
m ABR + m ABS + m PTQ = 180º
[From (i) and (ii)]
m RBS + m PTQ = 180º [Angle addition property]
m RBS + m RTS = 180º
[P - R - R and T - Q - S]
BSTR is cyclic
[If opposite angles of a quadrilateral
are supplementary then the
quadrilateral is cyclic]
TANGENT SECANT THEOREM
If an angle with its vertex on the circle whose one side touches the
circle and the other intersects the circle in two points, then the
measure of the angle is half the measure of its intercepted arc.
In the adjoining figure,
A
ABC has its vertex B on the
circle, line BC is tangent to circle
at B and ray BA is a secant.
X

m ABC =
1
m(arc AXB)
2
B
C
 NOTE 
If an angle,
(i) has its vertex on the circle.
(ii) one arm is a tangent, and
(iii) the other arm is a secant, then, we will term such an angle as a
Tangent secant angle.
106
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EXERCISE - 2.4 (TEXT BOOK PAGE NO. 70)
1.
D
In the adjoining figure,
seg AB and seg AD are chords
of the circle. C be a point on
tangent to the circle at point A.
If m (arc APB) = 80º and BAD = 30º,
then find (i) BAC (ii) m (arc BQD)
m BAC =
1
m(arc APB)
2

m BAC =
1
× 80
2

m BAC = 40º
Sol.
m BAD =
1
m (arc BQD)
2
Q
B
(2 marks)
P
A
C
[Tangent secant theorem]
[Inscribed angle theorem]
1
m (arc BQD)
2
 m (arc BQD) = 30 × 2

30 =
 m (arc BQD) = 60º
EXERCISE - 2.4 (TEXT BOOK PAGE NO. 70)
6.
In the adjoining figure,
points B and C lie on tangent to
the circle drawn at point A.
Chord AD  chord ED.
D
C
Q
R
1
A
E
m (arc AQD)
2
and m (arc DRE) = 84º then determine P
(i) DAC (ii) FDA (iii) FED (iv) BAF
(5 marks)
F
B
Construction : Draw seg EA.
Sol.
In DEA,
seg ED  seg AD
[Given]
 DEA  DAE
......(i) [Isosceles triangle theorem]
If m (arc EPF) =
m DEA =
But,
1
m (arc AD)
2
.....(ii)
[Inscribed angle theorem]
1
m DAE =
m (arc DRE) .....(iii)
2
1
m (arc AQD)
2

m (arc AQD)
But,
m (arc DE)

m (arc AQD)

1
m (arc DE)
2
= m (arc DRE)
= 84º
= 84º
=
m (arc EPF) =


1
m(arc AQD)
2
[From (i), (ii) and (iii)]
[Given]
[Given]
1
× 84
2
m (arc EPF) = 42º
m (arc EPF) =
S C H O O L S E C TI O N
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



m DAC =
1
m (arc AQD)
2

m DAC =
1
× 84
2

m DAC = 42º

[Tangent
secant theorem]

1
m (arc AF)
2
1
m FDA =
× 150
2
m FDA = 75º
(iii)
m FED =
1
m (arc DAF)
2
m FED =
1
[m(arc AQD) + m(arc AF)] [Arc addition property]
2

m FED =
1
(84 + 150)
2

m FED =
1
× 234
2

m FED = 117º
(iv)
m BAF =
1
m(arc AF)
2

m BAF =
1
× 150
2

m BAF = 75º


m (arc AQD) + m (arc DRE) + m (arc EPF) + m (arc AF) = 360º
[Measure of a circle is 360º]
84 + 84 + 42 + m (arc AF) = 360
210 + m (arc AF) = 360 – 210
m (arc AF) = 360 – 210
m (arc AF) = 150º
(i)
(ii)

EDUCARE LTD.
m FDA =
[Inscribed angle theorem]
[Inscribed angle theorem]
[Tangent secant theorem]
Segment of a circle : A secant divides the circular region into two
parts. Each part is called a segment of the circle.
Alternate segment : Each of the two segments formed by the secant of
a circle is called alternate segment in relation with the other.
Angle formed in a segment : An angle inscribed in the arc of a segment
is called an angle formed in that segment.
C
In the adjoining figure,
secant AB divides the circular region into
two segments R1 and R2.
R1 and R2 are alternate segments
in relation with each other.
ACB is inscribed in arc ACB of segment R2.
 ACB is an angle formed in segment R2.
108
R2
B
A
R1
X
S C H O O L S E C TI O N
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ANGLES IN ALTERNATE SEGMENT
If a line touches a circle and from the point
A
D
of contact a chord is drawn then the angles
which this chord makes with the given line
X
are equal respectively to the angles formed
in the corresponding alternate segments.
In the adjoining figure,
B
C
1
m (arc AXB) .....(i) [Tangent secant theorem]
m ABC =
2
1
m (arc AXB) ......(ii) [Inscribed angle theorem]
m ADB =
2
 m ABC = m ADB
[From (i) and (ii)]

ABC ADB
 NOTE 
If a tangent secant angle and an inscribed angle intercept the same arc
then they are congruent.
EXERCISE - 2.4 (TEXT BOOK PAGE NO. 70)
2.
If the chord AB of a circle is parallel to the tangent at C, then prove
that AC = BC.
(3 marks)
B
A
Proof :

Take a point D on the tangent at
C as shown in the figure.
seg AB || line CD.
[Given]
C
 On transversal AC,
D
BAC  ACD
........(i) [Converse of alternate angles test]
But, ACD  ABC
.......(ii) [Angles in alternate segment]
In ABC,
BAC  ABC
[From (i) and (ii)]
 seg AC  seg BC
[Converse of Isosceles triangle theorem]

AC = BC
EXERCISE - 2.4 (TEXT BOOK PAGE NO. 70)
3.
Suppose AB and AC are equal chords of a circle and a line parallel to the
tangent at A intersects the chords at D and E. Prove that AD = AE.
(4 marks)
Construction : Draw seg BC.
Proof :
Take points R and S on the tangent at A
C
B
as shown in the figure
line DE || line RS
[Given]
 On transversal AD,
E
D
EDA  DAR
[Converse of
•
alternate angles test]
A
S
R
 EDA  BAR
.......(i)
[ B - D - A]
BAR  BCA
.......(ii) [Angles in alternate segment]

S C H O O L S E C TI O N
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 EDA  BCA
......(iii)
Similarly, we can prove that
DEA  CBA
......(iv)
In ABC,
seg AB  seg AC
 BCA  CBA
........(v)
In DEA,
EDA  DEA
 seg AD  seg AE

AD = AE
EDUCARE LTD.
[From (i) and (ii)]
[Given]
[Isosceles triangle theorem]
[From (iii), (iv) and (v)]
[Converse of isosceles triangle theorem]
EXERCISE - 2.4 (TEXT BOOK PAGE NO. 70)
Suppose ABC is a triangle inscribed in a circle, the bisector of ABC
intersects the circle again in D, the tangent at D intersect the line BA
and line BC in E and F respectively. Prove that EDA  FDC.
Proof :
B
4.
••
A
 EDA
 FDC
But, ABD

EDA




ABD
CBD
CBD
FDC
E
........(i)
........(ii)
.......(iii)

C
D
F
[Angles in alternate segment]
[ Ray BD bisects ABC]
[From (i), (ii) and (iii)]
EXERCISE - 2.4 (TEXT BOOK PAGE NO. 70)
5.
Suppose two circles are touching at A. Through A two lines are drawn
intersecting one circle in P,Q and the other in X. Y. Prove that PQ || XY.
(4 marks)
Proof :
Two circles are touching at A.
M
P
therefore there are two possibilities
Y
(a) Two circles touch each other
externally at A.
A
(b) Two circles touch each other
internally at A.
X
Q
(a) If two circles touch each other
N
externally at A.
Draw a common tangent MN to
Proof :
the two circles through point A.
PQA  PAM
......(i)
[Angles in alternate segment]
XYA  XAN
......(ii)
.....(iii)
[Vertically opposite angles]
But, PAM  XAN
 PQA  XYA
[From (i), (ii) and (iii)]
 PQY  XYQ
[ Q - A - Y]
 seg PQ || seg XY
[By alternate angles test]

110
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(b) If two circles touch each other internally at A.
P

M
X
A
Construction : Draw a common tangent MN to
Y
the two circles through point A.
Q
Proof :
PQA  PAM
.......(i)
[Angles in
N
XYA  XAM
alternate segment]
 XYA  PAM
.......(ii)
[ P - X - A]
 PQA  XYA
[From (i) and (ii)]
 seg PQ || seg XY
[By corresponding angles test]
EXERCISE - 2.4 (TEXT BOOK PAGE NO. 71)
8.
Two circles intersect each
other at A and B. Let DC be a
common tangent touching the
circle C and D.
Prove that CAD +CBD = 180º.
Proof :
m BAC = m BCD ......(i)
A
(3 marks)
B
C
D
[Angles in alternate segments]
m BAD = m BDC ......(ii)
In BCD,
m BCD + m BDC + m CBD = 180º
[Sum of the measures of
angles of a triangle is 180º]
 m BAC + m BAD + m CBD = 180º
[From (i) and (ii)]
 m CAD + m CBD = 180º
[Angle addition property]
EXERCISE - 2.4 (TEXT BOOK PAGE NO. 71)
9.
Let the circles with centre P and
Q touch each other at point A.
Let the extended chord AB intersect
B
P
A
E
Q
the circle with centre P at point E
and the chord BC touches the circle
C
M
with centre P at the point D.Then prove
D
(4 marks)
that ray AD is an angle bisector of the CAE.
Proof :
In MAD,
MA = MD
[The lengths of two tangent
segments from an external point
to a circle are equal]
 m MAD = m MDA
[Isosceles triangle theorem]
Let,
........(i)
m MAD = m MDA = xº
CAM
 ABC
[Angles in alternate segments]
Let,
........(ii)
m CAM = m ABC = yº
[Angles Addition property]
m CAD = m CAM + m MAD
m CAD = (x + y)º
......(iii) [From (i) and (ii)]
S C H O O L S E C TI O N
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




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DAE is an exterior angle of ADB
m DAE = m ADB + mABD
[Remote Interior angles theorem]
m DAE = m ADM + m ABC
[D - M - C - B]
mDAE = (x + y)º
......(iv) [From (i) and (ii)]
m CAD = m DAE
[From (iii) and (iv)]
ray AD is an angle bisector of CAE.
EXERCISE - 2.4 (TEXT BOOK PAGE NO. 71)
7.
Let M be a point of contact of two internally touching circles. Let line
AMB be their common tangent. The chord CD of the bigger circle touches
the smaller circle at point N and chord CM and chord DM of bigger
circle intersect smaller circle at the points P and R respectively. Prove
that CMN  DMN.
(4 marks)
N
C
P
D
R
M
A
B
Construction : Draw seg NR.
Proof :
CMA
 CDM
[Angles in alternate segments]
Let,
m CMA = mCDM = xº
........(i)
NMA
 NRM
[Angles in alternate segments]
Let,
mNMA  NRM = yº
.......(ii)
m NMC = m NMA – m CMA
[Angle Addition property]
 m NMC = (y – x)º
......(iii) [From (i) and (ii)]
NMR  DNR
.......(iv) [Angles in alternate segment]
NRM is an exterior angle of  NDR
 m NRM = m NDR + mDNR
[Remote interior angles]
 m NRM = CDM + m DNR
[ C - N - D and D - R - M]
 y
= x + mDNR
[From (i) and (ii)]
 m DNR = (y – x)º
......(v)
 m NMR = (y – x)º
[From (iv) and (v)]
 m NMD = (y – x)º
......(vi) [D - R - M ]
 CMN  DMN
[From (iii) and (vi)]
ALTERNATIVE METHOD :
CMA  CDM
[Angles
in alternate segments]
Let,
m CMA = m CDM = xº
m NMA =
1
m (arc NM)
2
N
C
D
.....(i)
P
.....(ii)
A
R
M
1
.....(iii)
m CNM = m (arc NM)
2
 m NMA = m CNM
[From (ii) and (iii)]
Let,
m NMA = m CNM = yº
......(iv)
m CMN = m NMA – m CMA
[Angle addition property]
112
B
S C H O O L S E C TI O N
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 m CMN = (y – x)º
......(v) [From (i) and (iv)]
CNM is an exterior angle of NMD,
 m CNM = m NDM + m DMN
[Remote interior angles theorem]
 y = x + m DMN
[From (i), (vi) and C - N - D]
 m DMN = (y – x)º
......(vi)
 CMN  DMN
[From (v) and (vi)]
THEOREM
If two secants of a circle intersect inside or outside the circle then
the area of the rectangle formed by the two line segments
corresponding to one secant is equal in area to the rectangle formed
by the two line segments corresponding to the other.
In the adjoining figure,
A
D
chords AB and CD intersect each
other at point P inside the circle.
 OA × OB = OC × OD
O
C

B
A
In the adjoining figure,
chords AB and CD intersect each
other at point P outside the circle.
OA × OB = OC × OD
B
O
D
C
THEOREM
Statement : If a secant and a tangent of a circle intersect in
the circle then the area of the rectangle formed by the two
corresponding to the secant is equal to the area of the square
segment corresponding to the tangent.
Given :(i) line PAB is a secant intersecting
the circle at points A and B.
(ii) line PT is a tangent to the circle at point T.
To Prove: PA × PB = PT²
Construction: Draw seg BT and seg AT.
Proof :In PTA and PBT,
TPA   BPT
[Common angle]
B
PTA   PBT
[Angles in alternate segment]
 PTA ~ PBT
[By AA test of similarity]


PT
PA
=
PB
PT
PA × PB = PT²
a point outside
line segments
formed by line
(2 marks)
T
P
A
[Corresponding sides of similar triangles]
EXERCISE - 2.5 (TEXT BOOK PAGE NO. 75)
3.
Sol.
C
In the adjoining figure,
a tangent segment PA touching
a circle in A and a secant PBC
are shown. If AP = 15 and BP = 10,find BC.
Line PBC is a secant intersecting
the circle at points B and C and
line PA is a tangent to the circle at point A.
S C H O O L S E C TI O N
B
A
(2 marks)
P
113
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GEOMETRY



CP × BP
CP × 10
CP × 10

CP

CP
CP
22.5
BC
BC



= AP²
= (15)²
= 225
225
=
10
= 22.5 units
= BC + BP
= BC + 10
= 22.5 – 10
= 12.5 units
EDUCARE LTD.
[Tangent secant property]
[ C - B - P]
EXERCISE - 2.5 (TEXT BOOK PAGE NO. 75)
2.
Sol.
ABCD is a rectangle. Taking AD as a diameter, a semicircle AXD is drawn
which intersects the diagonal BD at X. If AB = 12 cm, AD = 9 cm then find
the values of BD and BX.
(2 marks)
9
A
D
In ABD
[Angle of a rectangle]
m BAD= 90º
BD2 = AB2 + AD2
[By pythagoras theorem]

BD2 = 122 + 92
[Given]
12
X

BD2 = 144 + 81

BD2 = 225
C

BD = 15 cm
[Taking square roots] B
m BAD = 90º
[Angle of a rectangle]
 line BA is a tangent to the circle at point A
[A line perpendicular to the radius at its
outer end is a tangent to the circle]
 Line AB is a tangent and line BXD is a secant intersecting at points X
and D

AB 2 = BX . BD
[Tangent secant property]

122 = BX . 15

144 = BX . 15
144

BX =
15

BX
= 9.6 cm
EXERCISE - 2.5 (TEXT BOOK PAGE NO. 75)
5.
In the adjoining figure,
A
two circles intersect each other
in two points A and B. Seg AB is
the chord of both circles. Point C
is the exterior point of both the
B
M
N
circles on the line AB. From the
point C tangents are drawn to the
circles touches at M and N.
C
Prove that CM = CN.
(2 marks)
Proof :
Line CBA is a secant intersecting the circle at points B
and A and line CM is a tangent to the circle at point M.
 CM² = CB × CA
.......(i)
[Tangent secant property]
Line CBA is a secant intersecting the circle at points B and A and
line CN is a tangent to the circle at point N.
 CN² = CB × CA
......(ii)
[Tangent secant property]
 CM² = CN²
[From (i) and (ii)]
 CM = CN
[Taking square roots]
114
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EXERCISE - 2.5 (TEXT BOOK PAGE NO. 75)
6.
In the adjoining figure, given two
concentric circles of radii 5 and 3,
D
find the length of a chord of larger
A
O
circle which touches the smaller one.
If BD = 5, find BC.
(3 marks)
E
C
Construction : Let the centre of the circle be O
Let seg AB touch the smaller circle at E.
B
Sol.
For the smaller circle
Line BE is a tangent to the circle at E and line CD is a secant
intersecting the circle at points C and D.
 BE2 = BC × BD
......(i) [Tangent secant property]
In OEB,
......(ii) [Radius is perpendicular to tangent]
m OEB = 90º
2
2
2
[By Pythagoras theorem]
OB = OE + BE

52 = 32 + BE2
[Given]
2

25 = 9 + BE
 25 – 9 = BE2

BE2 = 16

BE = 4 units
.....(iii) [Taking square roots]
For the larger circle,
[From (ii)]
seg OE  chord AB

AB = 2 × BE
[Perpendicular from the centre of the
circle to the chord bisects the chord]

AB = 2 × 4

AB = 8 units

(4) 2 = BC × 5
[From (i) and (iii)]

16 = BC × 5
16

BC =
5
 BC = 3.2 units
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 194)
Q
P
C
As shown in the adjoining figure, two circles
intersect each other in points A and B.
A
Two tangents touch these circles in
points P, Q and R, S as shown. Line AB
intersects seg PQ in C and seg Rs in D.
B
Show that C and D are midpoints of
S
R D
(2 marks)
seg PQ and RS respectively.
Proof :
line PC is tangent to the circle at point P and line CAB is a secant.
 PC 2 = CA × CB
.......(i) [Tangent secant property]
Line QC is a tangent to the circle at point C and line CAB is a secant.
 QC2 = CA × CB
......(ii) [Tangent secant property]
2
2
 PC = QC
[From (i) and (ii)]
 PC = QC
[Taking square roots]
 C is the midpoint of seg PQ.
Similarly, we can prove
D is the midpoint of seg RS.
9.
S C H O O L S E C TI O N
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EXERCISE - 2.5 (TEXT BOOK PAGE NO. 75)
4.
Point M, in the interior of the circle, is the point of intersection of two
chords AB and CD of the same circle. Show that CM × BD = BM × AC.
(2 marks)
D
Proof :
A
M
B
CAB  BDC
In CAM and BDM,
CAM  BDM
AMC  DMB
 CAM ~ BDM
CM
AC

=
BM
BD
 CM × BD = BM × AC
........(i)
[Angles inscribed C
in the same arc are congruent]
[From (i), A - M - B and D - M - C]
[Vertically opposite angles]
[By AA test of similarity]
[c.s.s.t.]
 Concyclic Points :
B
A
Given points lying on the same circle are
called concyclic points.
In the adjoining figure,
points A,B,C and D are concyclic
D
as all the points lie on the same circle.
C
 NOTE 
(i) Two distinct points are always concyclic.
(ii) Three points in a plane are concyclic if and only if they are non-collinear.

D
If A, B, C, D are four points such that the
points C, D are on the same side of the line
segment AB and line segment AB subtends
equal angles at C and D then A, B, C and D
are concyclic.
C
x
x
A
B
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 196)
In ABC, seg AB  seg AC and P is any point on AC. Through C, a line is
drawn to intersect BP produced in Q such that ABQ  ACQ. Prove
A
1
Q
that AQC = 90º +
BAC. (2 marks)
2
Proof :
P
23.




116
In ABC,
seg AB  seg AC
[Given]
C
B
ABC  ACB
......(i)
[Isosceles triangle theorem]
ABQ  ACQ
[Given]
seg AQ subtends congruent angles at points B and C which are on the
same side of line AQ.
Points A, B, C, Q are concyclic.
S C H O O L S E C TI O N
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EDUCARE LTD.


ABCQ is cyclic
m AQC + m ABC = 180º

m AQC = 180 – m ABC
In ABC,
m ABC + m ACB + m BAC = 180º






[By definition]
[ Opposite angles of a cyclic
quadrilateral are supplementary]
......(ii)
[ Sum of the measure of angles
of a triangle is 180º]
[From (i)]
m ABC + m ABC + m BAC = 180º
2m ABC + m BAC = 180
2m ABC = 180 – m BAC
1
m ABC = 90 –
m BAC .....(iii)
[Dividing throughout by 2]
2
1
m BAC) [From (ii) and (iii)]
m AQC = 180 – (90 –
2
1
m AQC = 180 – 90 + m BAC
2
1
m AQC = 90º +
m BAC
2
PROBLEM SET - 2 (TEXT BOOK PAGE NO. 196)
ABC is inscribed in a circle.
N P
A
Point P lies on a circumscribed circle of a
triangle such that through point P, PN,
M
PM and PL are perpendicular on sides of
triangle (possibly by increasing sides).
C (5 marks)
Prove that point N, M and L are collinear.
L
Construction : Draw seg NM, seg ML and seg AP.
Proof :
m PNA = 90º
......(i) [Given] B
m PMA = 90º
......(ii)
Adding (i) and (ii),
m PNA + m PMA = 90º + 90º
 m PNA + PMA = 180º
 PNAM is cyclic
[If opposite angles of a quadrilateral
are supplementary then the
quadrilateral is cyclic]
 Point P, N, A and M are concyclic.
 PAN  PMN
.....(iii) [Angles inscribed in same arc are congruent]
PABC is cyclic
[By definition]
 PAN  PCB
[Exterior angles of a cyclic quadrilateral
is equal to its interior opposite angle]
 PAN  PCL
.....(iv) [C - L - B]
 PMN  PCL
......(v) [From (iii) and (iv)]
PMC  PLC
[ each is 90º]
 seg PC subtends congruent angles at points M and L lying on the same
side of line PC
 Points P, M, L and C are concyclic
 PMLC is cyclic
[By definition]
 m PCL + mPML = 180º .....(vi)[Opposite angles of cyclic quadrilateral
are supplementary]
 m PMN + m PML = 180º
[From (v) and (vi)]
Also, PMN and PML are adjacent angles
 PMN and PNL form a linear pair [Converse of linear pair axiom]
 ray MN and ray ML are opposite rays
 Points N, M and L are collinear.
22.
S C H O O L S E C TI O N
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HOTS PROBLEM
(Problems for developing Higher Order Thinking Skill)
A
In the adjoining figure,
DE intersects the sides of ABC
P
in point P and Q such that
Arc AD = Arc DB and Arc AE = Arc EC. D
Show that PQC  BPQ.
(5 marks)
Construction : Draw seg DC and seg BF
B
Proof :
m (arc AD) = m (arc BD) ......(i) [Given]
9.
1
m (arc AD)
2
1
m BED =
m (arc BD)
2
 m ACD = m BED
 ACD  BED
 QCD  BEP
m (arc AE) = m (arc CE)
m ACD =
m ABE =
1
m (arc AE)
2
Q
E
C
......(ii) [Inscribed angle theorem]
.....(iii)
[From (i), (ii) and (iii)]
......(iv) [ A - Q - C and D - P - E]
.......(v) [Given]
....(vi)
[Inscribed angle theorem]
1
m (arc CE) .....(vii)
2
m ABE = m CDE
[From (v), (vi) and (vii)]
ABE  CDE
PBE  CDQ
......(viii) [ A - P - B and D - Q - E]
In BEP and DCQ,
BEP  QCD
[From (iv)]
PBE  CDQ
[From (viii)]
BEP ~ DCQ
[By AA test of similarity]
BPE  DQC
[c.a.s.t.]
BPQ  PQC
[ P - Q - E and D - P - Q]
m CDE =






10.
In ABC, A is an obtuse angle.
P is the circumcentre of ABC.
Prove that PBC = A – 90º.
(5 marks)

B
P
Construction : Extend seg BP to intersect the
D
circle at point D, B - P - D and
A
C
draw seg AD.
Proof :
m BAD = 90º
.......(i) [Angle subtended by a semicircle]
DBC  DAC
[Angles inscribed in same arc are
congruent]
i.e. PBC  DAC
......(ii) [B - P - D]
BAC = BAD + DAC
[Angle addition property]

 A = 90º + PBC
[From (i) and (ii)]

A – 90 = PBC

PBC = A – 90º
118
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11.
In the adjoining figure,
BC is a diameter of the circle
A
with centre M. PA is a tangent
at A from P which is a point on
line BC. AD  BC.
Prove that DP2 = BP × CP – BD × CD. B
P
D
C
M
(5 marks)
Construction : Draw seg AB and seg AC
Proof :
In ADP,
m ADP = 90º
[Given]
2
2
2
 AP = AD + DP
[By Pythagoras theorem]
 DP2 = AP2 – AD2
......(i)
Line AP is a tangent to the circle at point A and line PCB is a secant to
the circle at points C and B
 AP2 = BP × CP
......(ii)
In BAC,
m BAC = 90º
[Angle subtended by semicircle]
seg AD  hypotenuse BC
 AD2 = BD × CD
......(iii) [Property of geometric mean]
 DP2 = BP × CP – BD × CD
[From (i), (ii) and (iii)]
12.
ABC is an equilateral triangle.
Bisector of B intersects
circumcircle of ABC in point P.
Prove that CQ = CA.
(5 marks) C
B
A
P
Q
Proof :
ABC is an equilateral triangle [Given]
 m ABC = m BAC = m ACB = 60º
......(i)
[Angle of an equilateral triangle]
1
[ ray BP bisects ABC]
m CBP = m ABC
2
1
 m CBP =
× 60
[From (i)]
2
 m CBP = 30º
.....(ii)
CBP  CAP
....(iii) [Angles inscribed in the same
are congruent]
 m CAP = 30º
[From (ii) and (iii)]
 m CAQ = 30º
.....(iv) [ A - P - Q]
ACB is an exteror angle of CQA
 m ACB = m CQA + m CAQ
[Remote interior angle theorem]
 60 = m CQA + 30
[From (i) and (iv)]
 m CQA = 60 – 30
 m CQA = 30º
.....(v)
In CQA
CAQ  CQA
[From (iv) and (v)]
 seg CQ  seg CA
[Converse of Isosceles triangle theorem]
 CQ = CA
S C H O O L S E C TI O N
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13.
Two circles with centre O and
P intersects each other in poitn C
and D. Chord AB of the circle with
centre O touches the circle with
centre P in point E.
Prove that ADE + BCE = 180º.
Construction : Draw seg DC.
Proof :
ABCD is cyclic
m CBE = m ADC
......(i)
EDUCARE LTD.
D
O
P
C
B
A
E
(4 marks)
[By definition]
[Exterior angle property of cyclic
quadrilateral]
......(ii) [Angles in alternate segment]
m CDE = m CEB
In BCE,
m CBE + m CEB + m BCE = 180º
[Sum of the measures of the
angles of a triangle is 180º]
[From (i) and (ii)]
[Angle addtion property]
 m ADC + m CDE + m BCE = 180º
 m ADE + m BCE = 180º
58.
Show that the radius of incircle of right angle triangle is equal to the
difference of half of the perimeter and the hypotenuse.
(5 marks)
A

Given : (i)In ABC, m ABC = 90º
x
(ii) Circle with centre O touches the
x
sides AB, BC and AC at point L, M
N
and N respectively.
O
L
1
z
y
To prove : r =
(AB + BC + AC) – AC
2
C
B y
z
M
Proof :
Let the radius of the incircle be r
 OL = OM = ON = r
[Radii of the same circle]
AL = AN = x
.....(i) [Lengths of the two tangents
BL = BM = y
.....(ii) segment from an external point to
CM = CN = z
....(iii) a circle are equal]
AB + BC + AC = AL + BL + BM + CM + AN + CN
[A - L - B, B - M - C and A - N - C]
AB + BC + AC = x + y + y + z + x + z [From (i), (ii) and (iii)]
 AB + BC + AC = 2x + 2y + 2z
 AB + BC + AC = 2 (x + y + z)
1

(AB + BC + AC) = x + y + z
2
1

(AB + BC + AC) = (x + z) + y
2
1

(AB + BC + AC) = AC + y
[From (i), (iii) and A - N - C]
2
1

(AB + BC + AC) – AC = y ......(iv)
2
In LBMO,
m LBM = 90º
[Given]
m BLO = m BMO = 90º
[Radius is perpendicular to tangent]
 m LOM = 90º
[Remaining angle]
 LBMO is a rectangle
[By definition]
 LB = OM
[Opposite sides of an rectangle]
 LB = OM = y = r
......(v)
1

(AB + BC + AC) – AC = r
[From (iv) and (v)]
2
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If two circles are internally touching at point P. A line intersect those
two circles in point A, B, C, D respectively then prove that APB CPD.
M (5 marks)
A
B
P
C
D
N
Construction : Draw a common tangent MN at point P.
Proof :
APM  ADP
[Angles in alternate segment]
Let,
......(i)
m APM = m ADP = x
BPM  BCP
[Angles in alternate segment]
Let,
.....(ii)
m BPM = m BCP = y
[Angle addition proerty]
m APB = m BPM – m APM
 m APB = (y – x)
.....(iii) [From (i) and (ii)]
BCP is an exterior angle of CPD,
 m BCP = m CPD + m CDP
[Remote interior angles theorem]
 y = m CPD + x
[From (i) and A - C - D]
 m CPD = (y – x)
......(iv)
 m APB = m CPD
[From (iii) and (iv)]
 APB  CPD
MCQ’s
1.
P is the centre of the circle. T is a point on the circle and TB is a tangent.
PBT = 30º and radius is 10 cm. What is length of TB ?
(a) 20 cm
(b) 20 3 cm
(c) 10 cm
(d) 10 3 cm
2.
TP and TQ are tangents to the circle with centre O. If PTQ = 80º then
what is the measure of POT ?
(a) 100º
(b) 80º
(c) 50º
(d) 40º
3.
Two circles with centres P and Q touch externally at T. AB is a common
tangent to the circles. What is the measure of ATP ?
(a) 30º
(b) 45º
(c) 60º
(d) 90º
4.
Circles with centres A and B and radius 10 cm and 5 cm touch each other
internally at point C. P is any point on AC such that AP = 7 cm. How many
tangents can be drawn to the circles from point P ?
(a) 0
(b) 1
(c) 2
(d) 1 or 2
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5.
P is the centre of the circle. AB is a tangent to the circle at A. PAB is an
isosceles triangle. What is the measure of APB ?
(a) 30º
(b) 45º
(c) 60º
(d) 90º
6.
A circle is inscribed in ABC in which BC = 6 cm, AB = 8 cm and B = 90º.
What is the radius of the circle ?
(a) 4 cm
(b) 3 cm
(c) 2 cm
(d) 9 cm
ABC is inscirbed in a circle AB = 15 cm, AC = 8 cm, A = 90º. What is the
radius of the circle ?
(a) 8.5 cm
(b) 8 cm
(c) 7.5 cm
(d) 4 cm
7.
8.
Rectangle ABCD is inscribed in a circle with centre P and diameter 41 cm. If
one of the sides of the rectangle is 9 cm then what is the perimeter ?
(a) 49 cm
(b) 50 cm
(c) 98 cm
(d) 100 cm
9.
ABCD is cyclic quadrilateral such that 2 A = 3 C. What is the measure
of C ?
(a) 108º
(b) 72º
(c) 100º
(d) 80º
10.
In the adjoining figure,
TA and TB are tangents from T.
T
CD is also a tangent at P.
It TB = 12 cm. Find perimeter of TCD.
(a) 36 cm
(b) 18 cm
(c) 20 cm
(d) 24 cm
A
C
P
D
B
11.
seg AT is a tangent at T for a circle with centre O. TOB is the diameter. If
AOB = 130º then what is measure of OAT ?
(a) 40º
(b) 50º
(c) 65º
(d) 90º
12.
An angle substended by a semicircle at a point outside the circle is
.................. angle.
(a) right
(b) an acute
(c) an obtuse
(d) 180º
13.
A cyclic parallelogram is ................ .
(a) square
(b) rhombus
(c) rectangle
(d) kite
14.
If two secants AB and CD of a circle intersect in point ‘P’ then
AP CP

(a)
(b) AP = BP
BP PD
(c) CP = CD
(d) AP × BP = CP × DP
15.
If circles with centres A, B and C with radius 5 cm, 6 cm and 7 cm touch
each other externally. The perimeter of ABC is ............. cm.
(a) 18
(b) 23
(c) 36
(d) 40
122
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16.
Two concentric circles are of radii 5 cm and 3 cm. The length of the chord
of the larger circle which touches the smaller circle will be .................. .
(a) 8
(b) 4
(c) 6
(d) 10
17.
A chord divides a circle into 2 arcs measuring 2x and 7x. The measure of
the minor arc is .............. .
(a) 20
(b) 40
(c) 80
(d) 140
18.
seg AB and seg AD are chords
of the circle and AC is a tangent.
If m (arc APB) = 70º, m (arc BQD) = 16º,
then m CAD = ............. .
(a) 35
(c) 32
(b)
(d)
D
Q
B
•P
43
27
A
•
C
19.
Two circles with centres P and Q having diameter 25 cm and 15 cm
respectively touch each other externally at A, then the distance between P
and Q is ............. .
(a) 40 cm
(b) 10 cm
(c) 20 cm
(d) 45 cm
20.
seg AB is such that APB  AQB on the same side of seg AB. Hence,
points A, B, C, D are ............. .
(a) collinear
(b) coinciding
(c) concyclic
(d) non-concylic
21.
MT is a tangent and MSB is a secant MS = 4, MB = 16, hence MT = ?
(a) 8
(b) 10
(c) 12
(d) 30
22.
In the adjoining figure,
seg PT is a tangent and seg PB is secant.
 PTA ~ ............. .
(a) PAT
(c) PBT
(b)
(d)
P
BPT
TAB
A
B
T
23.
The diameter of two circles touching each other externally are 27 cm and
13 cm. Find the distance between their centre ............ .
(a) 20 cm
(b) 21 cm
(c) 19 cm
(d) 12 cm
24.
In a circle with centre P. Line QR is tangent at R. PQ = 13 cm, QR = 12 cm, then
radius PR = ......... cm.
(a) 25
(b) 5
(c) 10
(d) 1
25.
If TP and TQ are the two tangents to a circle with centre O. So that
POQ = 100º, then PTQ = ?
(a) 60
(b) 80
(c) 50
(d) 120
S C H O O L S E C TI O N
123
MT
GEOMETRY
EDUCARE LTD.
: ANSWERS :
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
(d)
(d)
(b)
(a)
(b)
(a)
(c)
(c)
(c)
(c)
(a)
(a)
(b)
10 3 cm
90º
45º
8.5 cm
72º
40º
rectangle
36
80
20 cm
8
20 cm
80
2.
4.
6.
8.
10.
12.
14.
16.
18.
20.
22.
24.
(a)
(a)
(c)
(c)
(d)
(b)
(d)
(a)
(b)
(c)
(b)
(b)
100º
0
12 cm
98 cm
24 cm
an acute
AP × BP = CP × DP
8
43
concyclic
PBT
5
1 Mark Sums
1.
In the adjoining figure,
if m ABC = 55º, then E
what is m AEB.
AEB  ABC
But, ABC = 55º
Sol.

A
•
C
B
[Angles in alternate segments]
[Given]
AEB = 55º
A
2.
What is the relation between
ABC and ADC of cyclic ABCD ?
Sol.
D


3.
B
C
ABCD is cyclic
m ABC + m ADC = 180º
[Given]
[Opposite angles of quadrilateral are
supplementary]
ABC and ADC are supplementary.
If seg AB is the diameter of the circle,
then find AB, if OB = 3 and OA = 4.
B
Sol.
O



A
seg AB is the diameter of the circle.
m AOB = 90º ......(i)
[Angle subtended by a semicircle]
In AOB,
m AOB = 90º
[From (i)]
AB2 = OA2 + OB2
[By Pythagoras theorem]
AB2 = 42 + 32
AB2 = 16 + 9
AB2 = 25

AB = 5 units

124
[Taking square roots]
S C H O O L S E C TI O N
MT
4.
GEOMETRY
EDUCARE LTD.
O is the centre of the circle.
If m ABC = 80º, the find
m (arc AC) and m (arc ABC).
B
O
C
A
m ABC
Sol.

80º

m (arc AC)
1
m (arc AC)
2
1
=
m (arc AC)
2
= 160º
=
[By Inscribed angle theorem]
m (arc ABC) = 360º – m (arc AC)
= 360 – 160

5.
Sol.
6.
m (arc ABC) = 200º
A
If PB = 3, PD = 4, PA = 6, find PC.
Chords AB and CD intersect each
P
other at point P outside the circle.
 PA × PB = PC × PD
 6 × 3 = PC × 4
63
 PC =
4
9
 PC =
2
 PC = 4.5 units
B
What is the relation between ABE
and ADC for cyclic ABCD ?
A
Sol.
D
•
E
ABE is an exterior angle of cyclic ABCD
C
ABE  ADC
[An exterior angle of a cyclic quadrilateral is
congruent to its interior opposite angle]
P is the centre of the circle and
line AB is a tangent at T. Radius
of the circle is 5 cm. Find the
distance of P from line AB.
A
Sol.
seg PT  line AB
PT = 5 cm

8.
C
B

7.
D
P
B
T
[Radius is perpendicular to the radius]
[Given]
Distance of P from line AB is 5 cm
Lines PM and PN are tangents to
P
the circle with centre O.
If PM = 7 cm, find PN.
Sol.
M
O
N
PM = PN
But, PM = 7 cm

[Length of the two tangent segments from an
external point to a circle are equal]
[Given]
PN = 7 cm
S C H O O L S E C TI O N
125
MT
GEOMETRY
9.
If m (arc PNQ) = 140º,
find m PQR.
Sol.
EDUCARE LTD.
P
N
R
Q
1
m (arc PNQ)
2
1
m PQR =
× 140º
2
m PQR =


10.
Sol.
11.
Sol.
Sol.
126
m PQR = 70º
Line AB is a tangent and line BCD
is a secant. If AB = 6 units, BC = 4 units,
find BD.
A
Line BCD is a secant intersecting
the circle at points C and D and
line BA is a tangent at A
C
 AB2 = BC × BD
B
2
 6 = 4 × BD
 36 = 4 × BD
36
 BD =
4
 BD = 9 units
D
Two circles with centres P and Q having diameter 25 cm and 15 cm
respectively touch each other externally at A. Find the distance between
P and Q.
Diameter of bigger circle = 25 cm
 Its radius (r1) = 12.5 cm
Diameter of smaller circle = 15 cm
 Its radius (r2) = 7.5 cm
The two circles with centres P and Q touch each other externally at A.
 Distance between P and Q = r1 + r2
= 12.5 + 7.5
= 20 cm

12.
[Tangent-secant theorem]
The distance between P and Q is 20 cm.
In the adjoining figure,
chords AB and CD intersect at E.
If DE = 6, BE = 3 and CE = 4, then
find AE.
Chords AB and CD intersect each
other at point E inside the circle
 AE × BE = CE × DE
 AE × 3 = 4 × 6
A
C
E
D
B
46
3

AE =

AE = 8 units
S C H O O L S E C TI O N
MT
13.
GEOMETRY
EDUCARE LTD.
In the adjoining figure,
m ABC = 57º. M is the centre
of the circle and line BC is a tangent.
seg BP is the diameter find,
(a) m (arc BQA) (b) m ABP.
Sol.
1
(a) m ABC =
m (arc BQA)
2
1
 57º =
m (arc BQA)
2
 m (arc BQA) = 114º
P
A
M
Q
B
C
[Tangent-secant theorem]
(b) seg PB is the diameter
 m (arc PAB) = 180º
[Measures of a semicircle]
 m (arc PA) + m (arc BQA) = m (arc PAB)
[Arc addition property]
 m (arc PA) + 114º = 180º
 m (arc PA) = 180º – 114º
 m (arc PA) = 66º
1
m (arc PA)
[Inscribed angle theorem]
m ABP =
2
1
 m ABP =
× 66º
2

14.
Sol.
O is the centre of a circle. TA is a tangent to the circle at point T. What
is measure of OTA ?
Line TA is a tangent to the circle at point T
[Given]

15.
Sol.
Sol.
m OTA = 90º
[Radius is perpendicular to the tangent]
Radius of the circle is 4 cm. What is the length of a chord of the circle ?
Radius of the circle is 4 cm
 Diameter of the circle = 4 × 2 = 8 cm
Diameter is the biggest chord of the circle

16.
m ABP = 33º
The length of the chord of the circle is 8 cm or less than 8 cm.
O is the centre of the circle. AB is the diameter. P is any point on the
circle other than A and B. What is m APB ?
seg AB is the diameter of the circle and P is the point on the circle

m APB = 90º
[Angle subtended by a semicircle]
17.
Sol.
What is the relation between the diameter and the radius of a circle ?
Diameter of a circle is twice its radius.
18.
In the adjoining figure,
line PQ and line PR are tangents
to the circle. So what is
P
m QPR + m QOR ?
Q
O
Sol.
m OQP = 90º
m ORP = 90º
S C H O O L S E C TI O N
R
......(i)
......(ii)
[Radius is perpendicular to the tangent]
127
MT
GEOMETRY
19.
Sol.


In OQPR,
m QPR + m QOR + m OQP + m ORP = 360º
[Sum of the measures of angles of a
quadrilateral is 360º]
m QPR + m QOR + 90º + 90º = 360º
[From (i) and (ii)]
m QPR + m QOR = 360º – 180º

m QPR + m QOR = 180º
O is the centre of the circle. AB is the longest chord of the circle. If
AB = 8.6 cm, what is the radius of the circle ?
AB is the longest chord of the circle
[Given]
But, diameter is the longest chord of the circle
 seg AB is the diameter of the circle
Diameter = 8.6 cm

20.
EDUCARE LTD.
Radius = 4.3 cm
In the adjoining figure,
if m APB = 30º, then m AOB ?
Sol.
P
O
B
A



1
m (arc AB) [Inscribed angle theorem]
2
1
30º =
m (arc AB)
2
m (arc AB) = 60º
m AOB = m (arc AB)
[Definition of measure of minor arc]
m APB
=
m AOB
= 60º

128
S C H O O L S E C TI O N
 MT EDUCARE PVT. LTD.
GEOMETRY
S.S.C.
Marks : 30
CHAPTER 2 : Circle
GEOMETRY
SET - A
Q.I. Solve the following :
(i) In the adjoining figure,
Q is the centre of circle and PM
and PN are tangent segments to
the circle. If MPN = 40º circle,
find MQN.
Duration : 1 hr. 15 min.
(4)
M
P
40º
Q
N
(ii)
If two circles with radii 8 and 3 respectively touch internally then
show that the distance between their centers is equal to the difference
of their radii, find that distance.
P
Q.II. Attempt the following :
(i) In the adjoining figure,
are four tangents to a circle at the
points A,B, C and D. These four tangents
A
form a parallelogram PQRS.
If PB = 5 and BQ = 3 then find PS.
B
5
B
18
M
O
(iii) Three congruent circles with
centres A, B and C and with
radius 5 cm each, touch each
other in points D, E, G as shown in
(a) What is the perimeter of ABC ?
(b) What is the length of side DE of DEF ?
MAHESH TUTORIALS PVT. LTD.
R
D
A
29
In the adjoining figure,
line AB is tangent to both the
circles touching at A and B.
OA = 29, BP = 18, OP = 61
then find AB.
(9)
Q
C
S
(ii)
3
P
61
D
A
B
E
F
C
7
 MT EDUCARE PVT. LTD.
GEOMETRY
Q.III. Solve the following :
(i) In the adjoining figure,
point A is a common point of contact l
of two externally touching circles
and line l is a common tangent
to both circles touching at B and C.
Line m is another common tangent
at A and it intersects BC at D.
Prove that (a) BAC = 90º
(b) Point D is the midpoint of seg BC.
(12)
m
B
D
C
A
Q
(ii)
In the adjoining figure,
points P, B and Q are
points of contact of the respective
tangents. line QA is parallel to line PC.
If QA = 7.2 cm, PC = 5 cm, find the
radius of the circle.
A
O
B
C
P
A
(iii) In a right angled triangle ABC, ACB = 90º a
circle is inscribed in the triangle with radius r.
a, b, c are the lengths of the sides BC, AC and
AB respectively. Prove that 2r = a + b – c.
other circle at A. Prove that the
points D, T and A are collinear.
P
C
Q.IV. Solve the following :
(i) In the adjoining figure,
points P and Q are the centers of
the circles. Radius QN = 3, PQ = 9.
M is the point of contact of the circles. N
Line ND is tangent to the larger circle.
Point C lies on the smaller circle.
Determine NC, ND and CD.
Q
O
B
R
(5)
D
C
Q
M
P
Best of Luck
8
MAHESH TUTORIALS PVT. LTD.
 MT EDUCARE PVT. LTD.
GEOMETRY
S.S.C.
Marks : 30
CHAPTER 2 : Circle
GEOMETRY
SET - B
Q.I. Solve the following :
(i) In the adjoining figure,
if m (arc APC) = 60º
and m BAC = 80º
Find (a) ABC (b) m (arc BQC)
Duration : 1 hr. 15 min.
B
C
80º
•P
A
(ii)
In the adjoining figure,
m (arc XAZ) = m (arx YBW).
Prove that XY || ZW
X
Y
A
B
Z
W
Q.II. Solve the following :
(i) Secants containing chords RS
and PQ of a circle intersects each
other in point A in the exterior of
a circle, as shown in figure If
m (arc PCR) = 26º and
m (arc QDS) = 48º
then find
(a) AQR (b) SPQ, (c) RAQ
(ii)
(9)
S
R
A
•D
C
P
Q
Two circles intersect each other
C
in points A and B. Secants through
A and B intersects circles in C, D and
M, N. Prove that CM || DN.
M
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(4)
Q
•
A
B
D
N
9
 MT EDUCARE PVT. LTD.
GEOMETRY
(iii) Find the radius of the circle passing through the vertices of a right
angled triangle when lengths of perpendicular sides are 6 and 8.
Q.III. Solve the following :
(i) ABCD is cyclic quadrilateral, lines AB and DC
intersect in the point F and lines AD and BC
intersect in the point E. Show that the
B
circumcircle of BCF and CDE intersect
in a point on the line EF.
F
(ii)
(12)
A
D
C
E
G
P
In a cyclic quadrilateral ABCD,
D
the bisectors of opposite angles A
and C meet the circle at P and Q
•
respectively. Prove that PQ is a
•
A
diameter of the circle.
× C
×
B
Q
(iii) In DABC, midpoints of sides AB,
AC and BC are P, Q and R respectively.
AS  BC. Prove that PQRS is a
cyclic quadrilateral.
P
B
A
Q
S
R
C
Q.IV. Solve the following :
(5)
(i) Prove that the quadrilateral formed by the angle bisectors of a cyclic
quadrilateral is also cyclic.
Best of Luck
10
MAHESH TUTORIALS PVT. LTD.
 MT EDUCARE PVT. LTD.
GEOMETRY
S.S.C.
Marks : 30
CHAPTER 2 : Circle
GEOMETRY
SET - C
Duration : 1 hr. 15 min.
Q.I. Solve the following :
(i) In the adjoining figure,
seg AB and seg AD are chords
of the circle. C be a point on
tangent to the circle at point A.
If m (arc APB) = 80º and BAD = 30º,
then find (i) BAC (ii) m (arc BQD)
(ii)
(4)
D
Q
B
P
A
C
C
In the adjoining figure,
a tangent segment PA touching
a circle in A and a secant PBC
are shown. If AP = 15 and BP = 10,find BC.
B
P
A
Q.II. Solve the following :
(i) In the adjoining figure,
two circles intersect each other
in two points A and B. Seg AB is
the chord of both circles. Point C
is the exterior point of both the
circles on the line AB. From the
point C tangents are drawn to the
circles touches at M and N.
Prove that CM = CN.
(ii)
(9)
A
N
C
Two circles intersects each
other at A and B. Let DC be a
common tangent touching the
circle C and D.
Prove that CAD +CBD = 180º.
A
B
C
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B
M
D
11
 MT EDUCARE PVT. LTD.
GEOMETRY
(iii) In the adjoining figure, given two
concentric circles of radii 5 and 3,
find the length of a chord of larger
circle which touches the smaller one. A
If BD = 5, find BC.
D
O
E
C
B
Q.III. Solve the following :
(12)
(i) Suppose AB and AC are equal chords of a circle and a line parallel to
the tangent at A intersects the chords at D and E. Prove that AD = AE.
(ii)
Let M be a point of contact of two
internally touching circles. Let line
AMB be their common tangent.
The chord CD of the bigger circle
C
touches the smaller circle at point N
and chord CM and chord DM of bigger
circle intersect smaller circle at the
points P and R respectively.
A
Prove that CMN  DMN
N
D
P
R
M
(iii) Let the circles with centre P and
Q touch each other at point A.
Let the extended chord AB intersect
the circle with centre P at point E
and the chord BC touches the circle
E
with centre P at the point D.Then prove
that ray AD is an angle bisector of the CAE.
B
P
A
M
Q
B
C
D
Q.IV. Solve the following :
(i) In the adjoining figure,
R
points B and C lie on tangent to
the circle drawn at point A.
E
Chord AD  chord ED.
1
m (arc AQD)
If m (arc EPF) =
P
2
and m (arc DRE) = 84º then determine
(a) DAC (b) FDA (c) FED (d) BAF
D
C
Q
(5)
A
F
B
Best of Luck
12
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