Download quintessence

INFOMATHS
OLD QUESTIONS-CW1
10. The number of distinct solutions (x, y) of the system
of equations x2 = y2 and (x – a)2 + y2 = 1 where ‘a’ is
any real number, can only be
NIMCET-2011
(a) 0, 1, 2, 3, 4 or 5
(b) 0, 1 or 3
(c) 0, 1, 2 or 4
(d) 0, 2, 3 or 4
11. Two circles x2 + y2 = 5 and x2 + y2 – 6x + 8 = 0 are
given. Then the equation of the circle through their
point of intersection and the point (1, 1) is
BHU-2011
(a) 7x2 + 7y2 – 18x + 4 = 0
(b) x2 + y2 – 3x + 1 = 0
(c) x2 + y2 – 4x + 2 = 0
(d) x2 + y2 – 5x + 3 = 0
12. The pair of straight lines joining the origin to the
common point of x2 + y2 = 4 and y = 3x + c
perpendicular if c2 is equal to
KIITEE-2010
(a) 20
(b) 13
(c) 1/5
(d) 1
13. Intercept on the line y = x by the circle x2 + y2 – 2x =
0 is AB. Equation of the circle on AB as diameter is
KIITEE-2010
(a) x2 + y2 + x – y = 0
(b) x2 + y2 – x + y = 0
(c) x2 + y2 + x + y = 0
(d) x2 + y2 – x – y = 0
14. The locus of a point which moves such that the
tangents from it to the two circles x2 + y2 – 5x – 3 =
0 and 3x2 + 3y2 + 2x + 4y – 6 = 0 are equal is
KIITEE-2010
(a) 2x2 + 2y2 + 7x + 4y – 3 = 0
(b) 17x + 4y + 3 = 0
(c) 4x2 + 4y2 – 3x + 4y – 9 = 0
(d) 13x – 4y + 15 = 0
15. If the circle 9x2 + 9y2 = 16 cuts the x-axis at (a, 0)
and (-a, 0), then a is
PGCET-2010
(a)  2/3 (b) 3/4 (c) 1/4 (d) 4/3
16. If y = x + c is a tangent to the circle x2 + y2 = 8, then
c is
(PGCET paper – 2009)
(a)  3
(b)  2
(c)  4
(d)  1
17. Two distinct chords drawn from the point (p, q) on
the circle x2 + y2 = px + qy, where pq  0 are
bisected by the x-axis then
(MCA : KIITEE - 2009)
(a) |p| = |q|
(b) p2 = 8q2
(c) p2 < 8q2
(d) p2 > 8q2
18. The equation of the circle having the chord x – y = 1
25
 0 as a diameter
of the circle x 2  y 2  x  3 y 
2
is
HYDERABAD CENTRAL UNIVERSITY - 2009
21
0
(a) x 2  y 2  3x  y 
2
25
0
(b) x 2  y 2  3x  y 
2
25
0
(c) x 2  y 2  x  3 y 
2
21
(d) x 2  y 2  3x  y   0
2
CIRCLES
1. The number of points (x, y) satisfying (i) 3x - 4y =
25 and (ii) x2 + y2  25 is
HCU-2012
(a) 0
(b) 1
(c) 2
(d) infinite
2. A circle and a square have the same perimeter. Then
HCU-2012
(a) their areas are equal
(b) the area of the circle is larger
(c) the area of the square is larger
(d) the area of the circle is  times the area of the
square
3. The lines 2x – 3y = 5 and 3x – 4y = 7 are the
diameters of a circle of area 154 square units. Then
the equation of this circle is (= 22/7)
PU CHD-2012
(A) x2 + y2 + 2x – 2y = 62
(B) x2 + y2 + 2x – 2y = 47
(C) x2 + y2 – 2x + 2y = 47
(D) x2 + y2 – 2x + 2y = 62
4. If a given point is P(10,10) and the Eq. of circle is
(x – 1)2 + (y – 2)2 = 144. Where does the pt. lies
Pune-2012
(a) inside
(b) on
(c) outside
(d) None of these
5. If the circles x2 + y2 + 2x + 2ky + 6 = 0 and x2 + y2 +
2ky + k = 0 intersect orthogonally, then k is
NIMCET-2012
3
3
(a) 2 of 
(b) – 2 or 
2
2
3
3
(c) 2 or
(d) – 2 or
2
2
6. The equation of circle passing through (-1, 2) and
concentric with x2 + y2 – 2x – 4y – 4 = 0 is :
BHU-2012
(a) x2 + y2 – 2x – 4y + 1 = 0
(b) x2 + y2 – 2x – 4y + 2 = 0
(c) x2 + y2 – 2x – 4y + 4 = 0
(d) x2 + y2 – 2x – 4y + 8 = 0
7. The radius of the circle on which the four points of
intersection of the lines (2x – y + 1) (x – 2y + 3) = 0
with the axes lie, is :
BHU-2012
5
5
5
(a) 5
(b)
(c)
(d)
2
2 2
4 2
8. A curve given in polar form as r = a(cos() + sec ())
can be written in Cartesian form as
HCU-2011
(a) x(x2 + y2) = a(2x2 + y2)
1 

(b) y 2  a  x 2  2 
x


1

(c) y  a  x  
x

(d) y = atan  + x
9. If 2x + 3y – 6 = 0 and 9x+ 6y – 18 = 0 cuts the axes
in concyclic points, then the center of the circle is:
NIMCET-2011
(a) (2, 3) (b) (3, 2) (c) (5, 5) (d) (5/2, 5/2)
1
INFOMATHS/MCA/MATHS/
INFOMATHS
19. Lines are drawn through the point P (-2, -3) to meet
the circle x2 + y2 – 2x – 10y + 1 = 0. The length of
the line segment PA, A being the point on the circle
where the line meets the circle is.
KIITEE - 2008 29.
(a) 4 3 (b) 16
(c) 48
(d) None
20. If the common chord of the circles x2 + (y - )2 = 16
and x2 + y2 = 16 subtend a right angle at the origin
30.
then  is equal to.
MCA : KIITEE - 2008
(a)  4 2 (b) 4 2 (c) 4
(d) 8
21. The equation of any tangent to the circle x2 + y2 – 2x
+ 4y – 4 = 0 is
31.
KIITEE - 2008
(a) y  mx  3 1  m 2
(b) y  mx  2  3 1  m 2
(c) y  mx  m  2  3 1  m 2
(d) None
22. The equation of the circle whose two diameters are
2x – 3y + 12 = 0 and x + 4y – 5 = 0 and the area of
22 

which is 154 sq. units, will be :    
7 

MP COMBINED - 2008
(a) x2 + y2 + 6x – 4y + 36 = 0
(b) x2 + y2 + 3x – 2y + 18 = 0
(c) x2 + y2 – 6x + 4y + 36 = 0
(d) x2 + y2 + 6x – 4y – 36 = 0
23. The circle x2 + y2 – 2x + 2y + 1 = 0 touches:
MP COMBINED - 2008
(a) Only x-axis
(b) Only y-axis
(c) Both the axes
(d) None of the axes
24. If the line hx + ky = 1 touches the circle
1
( x 2  y 2 )  2 , then the locus of the point (h, k)
a
will be:
MP COMBINED - 2008
(a) x2 + y2 = a2
(b) x2 + y2 = 2a2
25.
26.
27.
28.
32.
33.
34.
35.
(a) 2g1g2 + 2f1f2 = c1 + c2
(b) g1g2 + f1f2 = c1 + c2
(c) g1g2 + f1f2 = 2(c1 + c2)
(d) g1g2 + f1f2 + c1 + c2 = 0
For the given equation x2 + y2 – 4x + 6y – 12 = 0, the
centre of the circle is
KARNATAKA - 2007
(a) (-2, 3) (b) (-3, 2) (c) (3, - 2) (d) (2, - 3)
The circumference of the circle x2 + y2 + 2x + 6y –
12 = 0 is
KARNATAKA - 2007
(a) 2
(b) 8
(c) 3
(d) None
The equation of the circle passing through the origin
and making intercepts of 4 and 3 or OX and OY
respectively is ICET – 2005
(a) x2 + y2 – 3x – 4y = 0
(b) x2 + y2 + 4x + 3y = 0
(c) x2 + y2 + 3x + 4y = 0
(d) x2 + y2 – 4x – 3y = 0
The radius of the circle
16x2 + 16y2 = 8x + 32y – 257 = 0
Karnataka PG-CET : Paper 2006
(a) 8
(b) 6
(c) 15
(d) None of these
If two circles x2 + y2 + 2g1x + 2f1y + c1 = 0 & x2 + y2
+ 2g2x + 2f2y + c2 = 0 will cut each other and
c c
satisfies the relation g1 g 2  f1 f 2  1 2 . Then
2
angle between the circles will be
MP : MCA Paper – 2003
(a) π/3
(b) π/2
(c) 3π/2 (d) π/4
Two circles x2 + y2 + 2gx + c = 0 and x2 + y2 + 2fy +
c = 0 touch each other, then :
MP :– 2003
(a) g2 + f2 = c3
(b) g2 + f2 = c
2
2
2 2
(c) c(g + f ) = g f
(d) g2 + f2c = g2f2
S1 = x2 + y2 – 4x – 6y + 10 = 0
S2 = x2 + y2 – 2x – 2y – 4 Angle between S1 and S2 is
UPMCAT : paper – 2002
(a) 90
(b) 60
(c) 45
(d) None of these
The tangents of the circle x2 + y2 = 4 at the points A
and B meet at P(-4, 0). The area of the quadrilateral
PAOB where O is the origin is.
KIITEE - 2008
36.
2
a
(c) x2 + y2 = 1
(d) x 2  y 2 
2
Equation of the circle concentric to the circle x2 + y2
– x + 2y + 7 = 0 and passing through (-1, -2) will be:
(a) 4
(b) 6 2 (c) 4
(d) None
MP COMBINED - 2008 37. The limiting points of the system of coaxial circles
(a) x2 + y2 + x + 2y = 0
of which two members are x2 + y2 + 2x + 4y + 7 = 0
(b) x2 + y2 – x + 2y + 2 = 0
and x2 + y2 + 5x + y + 4 = 0 is:
(c) 2(x2 + y2) – x + 2y = 0
MP : MCA Paper – 2003
(d) x2 + y2 – x + 2y – 2 = 0
(a) (-2, 1) and (0, - 3)
(b) (2, 1) and (0, 3)
For the circle x2 + y2 – 4x + 2y + 6 = 0, the equation
(c) (4, 1) and (0, 6)
(d) None of these
of the diameter passing through the origin is:
38. The length of common chord of the circles (x – a)2 +
MP COMBINED - 2008
y2 = a2 and x2 + (y – b)2 = b2 is :
(a) x – 2y = 0
(b) x + 2y = 0
MP : MCA Paper – 2003
(c) 2x – y = 0
(d) 2x + y = 0
ab
2
2
(a) 2 a  b
(b)
The circle x2 + y2 + 2ax – a2 = 0:
2
a  b2
(MP COMBINED – 2008)
2ab
(a) touches x – axis
(c)
(d) None of these
(b) touches y – axis
a 2  b2
(c) touches both the axis
39. The locus midpoint of a chord of the circle x2 + y2 =
(d) intersects both the axes
4, which subtend angle 90 at the centre.
The circles x2 + y2 + 2g1x + f1y + c1 = 0 and
UPMCAT : paper – 2002
2
2
x + y + g2x + 2f2y + c2 = 0 cut each other
(a)
x
+
y
+
3
=
0
(b)
x2 + y2 = 0
orthogonally, then :
(c) x + y + 2 = 0
(d) x2 + y2 = 2
(MP COMBINED – 2008)
2
INFOMATHS/MCA/MATHS/
INFOMATHS
1+4+2–8+k=0
k=1
 Circle will be x2 + y2 – 2x – 4y + 1 = 0
CIRCLES
1
B
11
A
21
C
31
D
2
B
12
D
22
D
32
D
3
C
13
D
23
C
33
B
4
C
14
B
24
A
34
C
5
A
15
D
25
D
35
D
6
A
16
C
26
B
36
C
7
C
17
D
27
D
37
D
8
A
18
D
28
A
38
C
9
D
19
A
29
D
39
D
10
D
20
A
30
D
7.
y = 0, x  
4.
5.
6.
y = 0, x = - 3
2
1
2

Also  h    h 2   h  3  k 2
2

7
 h
4
 7 5
So centre is   , 
 4 2
 7 5
So c    ,  and any point. (-3, 0)
 4 2

Area of square = a2. Sq. units.
4
Clearly as  1

Hence Area of circle is more than area of a square.
3.
1
2
 1 
Let (h, k) be centre and (0, 1), (0, 3/2)   ,0  , (-3,
 2 
0) lie on circle  h2 + (k – 1)2 = h2 + (k – 3/2)2
 k = 5/2
Ans. (b) Let radius of circle be ‘r’ units
circumference = 2r units and length of square as ‘a’
units. So, perimeter of square = 4a units.
As 2r = 4a
2a
r
units

Area of circle = r2
 .4a 2
4a 2
sq. units.


2

x – 2y + 3 = 0
3
x = 0, y 
2
x = 0, y = 1
SOLUTIONS CIRCLES
1. Ans. (b) The given straight line 3x – 4y – 25 = 0
touches the circle x2 + y2  25 as d = r = 5.
Hence, there is only one solution or single point (x,
y) satisfying both equations.
2.
Ans. (c)
2x – y + 1 = 0
2
 7
 5
r     3   
 4
 4

Ans. (c) As, diameters are given.
 POI of diameters is the centre of the circle.
i.e. 2x – 3y = 5
3x – 4y = 7
Solving both equations, centre will be (1, - 1)
Area = 154
r2 = 154.
r = 7 units.
Equation of circle will be
(x – 1)2 + (y + 1)2 = 72
x2 + y2 – 2x + 2y – 47 = 0.
8.
2
2
50
 5  25
   

16
 4  16
5 2
5

4
2 2
Ans. (a)
r = a [cos  + sec ]
1 

r  a cos  
cos
 

r cos  = a [cos2 + 1]
relationship between Cartesian and polar coordinates
r cos  = x
r sin  = y
Ans. (c) As S = x2 + y2 – 2x – 4y – 139 = 0
S1 = (10)2 + (10)2 – 2(10) – 4(10) – 139
= 200 – 199 = 1 > 0
 Point P(10, 10) lies outside.
 tan  
y
x
x  a cos 2   1
 1

 a  2  1
 sec  
1


 a
 1
2
1

tan



Ans. (a) As both circles are orthogonal.
 2 g1 g2  f1 f 2   c1  c2
2[0 + k2] = 6 + k
2k2 = 6 + k
2k2 = 6 + k
2k2 – k – 6 = 0
2k2 – 4k + 3k – 6 = 0
2k (k – 2) + 3(k – 2) = 0
(k – 2) (2k + 3) = 0
3
k = 2, 
2
 x2

x  a 2
 1
2
x  y

x  x 2  y 2   a  2 x 2  y 2 
9.
Ans. (a) As, the circle is concentric with x2 + y2 – 2x
– 4y – 4 = 0
 Equation of Required circle will be
x2 + y2 – 2x – 4y + k = 0
As, it passes through (-1, 2)
3
Ans. (d) As the lines intersects the Y-axis at (3, 0)
and (2, 0)
And point where the lines intersects the Y axis are
(0, 3) and (0, 2) is
Now, the equation of circle passing through (0, 2) (0,
3) and (2, 0) is
INFOMATHS/MCA/MATHS/
INFOMATHS
x2  y 2
x
y 1
4
0 2 1
9
0 3 1
0
4
2 0 1
Expanding along C2:4 2 1
x2  y 2
x 9
3 1 2
4 0 1
y 1
4
2 1 0
9
3 1
No. of solutions = 2
Case iii
On Expanding along C3:x2 + y2 – 5x – 5y + 6 = 0
5 5
Centre of circle is  , 
2 2
10. Ans. (d) For the solutions of the system of equations,
we need to solve the equation simultaneously.
i.e x2 = y2 and (x – a)2 + y2 = 1
No. of solutions = 3
 (x – a)2 + x2 = 1
Case iv
2x2 – 2ax + a2 – 1 = 0
Disc. = (-2a)2 – 4.2(a2 – 1)
= 4a2 – 8a2 + 8 = 8 – 4a2
if 8 – 4a2 < 0
There will be no real value of x
 No solution exists. So, No. of sol = 0
Case ii if 8 – 4a2 = 0
There will be only one distinct value of x, 2 distinct
values of y correspondingly
So, no. of solutions = 2
Case iii if 8 – 4a2 > 0 and a perfect square
No. of solutions = 4
i.e. if a = 1
D=4
11. Ans. (a) Eq. of circle through POI of S1 = 0 and S2 =
So, equation becomes
2
0 is S1 + S2 = 0
2x – 2x = 0
The circle passing through Poi of x2 + y2 – 6x + 8
2x(x – 1) = 0
= 0 and x2 + y2 = 5 = 0 is x2 + y2 – 6x + 8 + (x2 + y2
x = 0, 1
– 5) = 0
 For x = 0, we have y = 0,
(1 + )x2 + (1 + )y2 – 6x + (8 - 5) = 0
And for x = 1, y =  1
Also, this circle passes through (1, 1)
Hence, no. of solutions = 3
2(1 + ) – 6 + 8 - 5 = 0
2
Case iv if 8 – 4a > 0, but not a perfect square we
-3 + 4 = 0
have 2 distinct values of x
4
Now, correspondingly we have 4 values of y.

So, no. of sol. = 4
3
 in all total possible solutions of the equations are
4
Putting   in Eq. (*)
0, 2, 3, 4.
3
Graphically, the possible graphs for x2 = y2 is set of 2
 Required equation of circle will be
lines passing through origin.
7 2 7 2
4
And, (x – a)2 + y2 = 1 is a circle with centre (a, 0)
x  y  6x   0
3
3
3
and radius = 1 unit
7x2 + 7y2 – 18x + 4 = 0
Case i
12. Ans. (d)

No. of solutions = 0.
2
 sin 45 
1
2
Case ii
4
INFOMATHS/MCA/MATHS/
INFOMATHS
x2 – px + 2q2 = 0
As there are two distinct chords drawn from A.
 Disc. > 0
p2 – 8q2 > 0
p2 > 8q2
p 2

3 0  o  c
32  12
 2
 c  20  c 2  20
18. Ans. (d)
13. Ans. (d) For the POI of straight line y = x and x2 +
y2 – 2x = 0
 x2 + x2 – 2x = 0  2x2 – 2x = 0
2x (x – 1) = 0
x = 0, 1
 y = 0, 1
Hence POI are (0, 0) and (1, 1)
Now, the equation of a circle with co-ordinates of
diameter as (0, 0) and (1, 1) will be
(x – 0) (x – 1) + (y – 0) (y – 1) = 0
x.(x – 1) + y. (y – 1) = 0
x2 + y2 – x – y = 0
In this case element of common chord S1 – S2 = 0 is
x–y=1
So from choices (d) choice
21
x 2  y 2  3x  y 
2
25 
 2
2
  x  y  x  3y    0
2

 - 2x + 2y + 2 = 0
Also x – y – 1 = 0
14. Ans. (b) Locus of a point which moves such that
tangents from it to the two circles S1 = 0 and S2 = 0
are equal will be S1 – S2 = 0, here multiple first circle
by 3 to make coeff. of x2, y2 same of S1 and S2 = 0
i.e. 3(x2 + y2 – 5x – 3) – (3x2 + 3y2 + 2x + 4y – 6)=0
-15x – 9 – 2x – 4y + 6 = 0
-17x – 4y – 3 = 0
17x + 4y + 3 = 0
19. Ans. (a) Length of tangent = PA  48  4 3
units
20. Ans. (a) Equation of common chord S1 – S2 = 0
2 y   2  0  y 

0
2
p = distance of (0, 0) from common chord

0
2  
2
1
16
15. Ans. (d) As the circle x  y 
9
Intersects the X-axis at (a, 0) and (-a, 0)
4
then a  
3
2
2
16. Ans. (c) As the line y = x + c
is tangent to circle x2 + y2 = 8
 condition of tangency will be satisfied.
c2 = a2(m2 + 1)
c2 = 8(1 + 1)
c2 = 16
c=4
Also


2
p
1
 cos 45 
 p2 2
4
2
 2 2    4 2
21. Ans. (c)
x2 + y2 – 2x + 4y – 4 = 0
(x – 1)2 + (y + 2)2 = 9 = 32
So equation of tangent
17. Ans. (d)
y + 2 = m(x – 1) a 1  m2
y  2  m  x  1  3 1  m2
y  mx  m  2  3 1  m 2
22. Ans. (d) As, the Poi of diameters
2x – 3y + 12 = 0 and x + 4y – 5 = 0
will be the centre of circle.
i.e. 2x – 3y + 12 = 0
4
x + 4y – 5 = 0
3
8x – 12y + 48 = 0
3x + 12y – 15 = 0
On adding 11x + 33 = 0
x = - 3 and y = 2
centre is (-3, 2) and area of circle = 154.
r2 = 154
22 2
r  154
7
As M is midpoint of AB.
yq

0
2
y=-q
 B(x, y) is (x, - q)
Now, B(x, - q) lines on the circle x2 + y2 – px – qy=0
 x2 + q2 – px + q2 = 0
5
INFOMATHS/MCA/MATHS/
INFOMATHS
154  7
=77
22
2
r = 49
 r = 7 units.
Hence, Equation of circle will be (x+3)2+(y–2)2 =72
(x2 + 6x + 9) + (y2 – 4y + n) = 49
x2 + y2 + 6x – 4y – 36 = 0
 Circumference  2 22 units.
r2 
31. Ans. (d) Let Equation of circle passing through
origin be
x2 + y2 + 2gx + 2fy = 0
Length of x-intercept = 2g units
 2g = 4.
Length of x-intercept = 2f units
 2f = 3
So, equation of circle will be
x2 + y2 + 4x + 3y = 0
23. Ans. (c) As the centre of circle
x2 + y2 – 2x + 2y + 1 = 0 is (1, -1)
and radius 
1
2
  1  1 = 1
2
 the circle will touch both the co-ordinate Axes.
32. Ans. (d) The circle is 16x2 + 16y2 – 8x + 32y – 257
=0
24. Ans. (a) As the line hx + ky = 1 touches the circle
1
257
x2  y 2  x  2 y 
0
1
2
2
2
16
x y  2
a
1

Centre is  , 1
 it will satisfy the condition of tangency of circle
4


as c2 = r2 (1 + m2) … * Eq.
2
257
1  16  257
2
1
1
And radius      1 

here r  and for hx + ky = 1
16
16
a
4
h
1
274
y
x
units.

k
k
16
h
1
m
, c
k
k
c c
33. Ans. (b) If the condition g1 g 2  f1 f 2  1 2 is
2
2
1
1  h 
* Eq. gives 2  2 1  2 
satisfied by the 2 given circles, then angle between
k
a  k 

a2 = h2 + k2 replacing (h, k) by (x, y)
the circles will be . As, the circles are orthogonal.
2
2
2
2
 locus will be x + y = a
34. Ans. (c) Centre1 (-g, 0) r1  g 2  c
25. Ans. (d) Equation of circle concentric with
x2 + y2 – x + 2y + 7 = 0 is
x2 + y2 – x + 2y + k = 0
As, it passes through (-1, -2)
 (-1)2 + (-2)2 – (-1) + 2 (-2) + k = 0
1+4+1–4+k=0
k=-2
So, the required circle is
x2 + y2 – x + 2y – 2 = 0
Centre2 (0, - f) r2  f 2  c
If the circles are touching each other. Then r1 + r2=d.
g2  c  f 2  c  g2  f 2
Squaring both sides:-
(g2 – c) + (f2 – c) = g2 + f2 2 g 2  c f 2  c in a
single line.
2 g 2  c f 2  c  2c
Squaring Again:(g2 – c) (f2 – c) = c2
g2f2 – c(f2 + g2) + c2 = c2
g2f2 = c(g2 + f2)
26. Ans. (b) Diameter of circle
x2 + y2 – 4x + 2y + 6 = 0
will pass through centre of circle.
Here x + 2y = 0 is the only equation passing through
centre (2, - 1)
So, x + 2y = 0 is the equation of the diameter.
35. Ans. (d) Centre1 (2, 3) radius1  3 units
Centre2 (1, 1) radius2  6
Then Angle between the circles is
2
2
and radius   a    0   a 2  2a2  a 2
r 2  r22  d 2 3  6  5
cos   1

So, the circle will interested both the co-ordinate
2r1r2
2 18
axes as its radius is greater than ‘a’ units.
4
2
2



28. Ans. (a) The given circles will intersect orthogonally
3
2 18 3 2
if 2[g1g2 + f1f2] = c1 + c2.
 2
  cos 1 

29. Ans. (d) The centre of the circle x2 + y2 – 4x+6y–12
 3 
is (2, - 3)
36. Ans. (c)
30. Ans. (d) For the circle, x2 + y2 + 2x + 6y – 12 = 0
Centre is (-1, - 3)
27. Ans. (d) The circle x2 + y2 + 2ax – a2 = 0
Has centre as (-a, 0)
And radius 
 1
2
  3  12   22
2
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INFOMATHS/MCA/MATHS/
INFOMATHS
POA  POB
 Area PAOB = 2 Area POA
Now, As PAO is a right angled .
with PA = Base,
AO = 2 units
PO = 4 units
PO2 = PA2 + AO2
(4)2 = PA2 + (2)2
 PA2 = 12
PA  12 units.
1
Area  POA 
12  2  12 sq. units
2
 Area PAOB  2 12  4 3 sq. units
PA  a units, PB  b units.
Eq. of radical Axis PQ is S – S' = 0
(x – a)2 + y2 – a2 – [x2 + (y – b)2 – b2] = 0
(x2 + a2 – 2ax + y2 – a2) – (x2 + y2 + b2 – 2by – b2)=0
-2ax + 2by = 0
ax – by = 0
b2
Now, BM 
a 2  b2
PM2 = PB2 – BM2  b2 
37. Ans. (d) System of coaxial system of S1 = 0, S2 = 0,
is S1 +  (S2 – S1) = 0
 (x2 + y2 + 2x + 4y + 7) +  (3x – y – 1) = 0
 x2 + y2 + (2 + ) x + (4 - )y + (7 - ) = 0
 Radius of coaxial system
 2
PM 
ab
a  b2
2
So, length of common chord PQ 
 4 
 
 
  7     0
 2   2 
 ( + 2)2 + (4 - )2 – 4 (7 - ) = 0
 22 = 8 = 0  2 = 4   =  2
    2  4    
 centre of coaxial  
,

2
2 

 =  2   = 2  (-2, ),  = - 2  (0, - 3)
2
a 2b2
b4

a 2  b2
a 2  b2
2
2ab
a 2  b2
39. Ans. (d)
OM
1
 cos 45 
OA
2
OA
2
2
OM 
 2   x  0   y  0  2
2
 x2 + y2 = 2
38. Ans. (c)
7
INFOMATHS/MCA/MATHS/