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Circle Geometry Review Solutions
1. d 
 1  32  10  6) 2
 272  4 17  16.49
 3  (1)  6  10 
Midpo int : 
,
  (1, 2)
2 
 2
2.
8 x
 3, x  14
2
m AB 
10  (6)
4

 1
1 3
4
5 y
1
  , y  6  (14,6)
2
2
7
 3 1
 1 7
 3 
midpo int   ,   y       x       y  x  3
3
 2 2
 2 3
 2 
7
In order for (3,4) to be the center it must be on the line. Check y  (3)  3  4  could be center
3
3. m AB 
3
7
,  slope 
7
3
 6  7 5  (2) 
4. Need midpoint (center) to determine the length of the radius: 
,
  (6.5,1.5)
2 
 2
r  (6.5  6) 2  (1.5  5) 2  12.5
d  (6.5  3) 2  (1.5  (1)) 2  18.5
Since the distance from the center to the point (3, -1) is greater than the radius, the point is outside circle.
5. a) Remember that distance is measured perpendicularly to the chord and perpendicular bisector of
chord passes through center.
r 2  32  4 2  r  5  diameter  10cm
b) Think – the two chords could be on the same side of center or on opposite sides of center!
Let distance of one chord be a, let distance of other chord be b:
132  a 2  12 2  a  5
132  b 2  5 2  b  12
The chords could be: 12  5  17cm apart or 12  5  7cm apart
8
6. A sketch is always a good idea!
B(-2, 2)
  2 4 28
midpt AB : 
,
  (1,3)
2 
 2
 12  4 6  8 
midpt AC : 
,
  (8,1)
2 
 2
8 1
7
m segment :

1 3 2
12  2 14 7
m BC :


62
4 2
y
6
C(12, 6)
4
2
x
5
−2
−4
10
(8, -1)
(1, -3)
−6
−8
A (4, -8)
−10
Segment and BC have same slope, therefore parallel.
Measure of segment is half measure of BC.
d segment  (8  1) 2  (1  3) 2  53 d BC  (12  2) 2  (6  2) 2  212  2 53
15
7.
6
4
2
R(-6, 5)
−12
−10
−8
−6
−4
−2
U(-10, -7)
  10  12  7  1 
midpt SU : 
,
  (1,4)
2
2 

y
S(12, -1)
2
−2
−4
−6
−8
−10
−12
−14
−16
4
6
8
10
12
x
14
T(8, -13)
  6  8 5  13 
midpt RT : 
,
  (1,4)
2 
 2
The diagonals have the same midpoint, therefore they bisect each other.
8. You are told that triangle ABC is isosceles so you do not have to prove that. Let D be midpoint of AB,
E be midpoint of BC and F be midpoint of AC. Now prove triangle DEF is isosceles . . .
y
1 5 7  3 
D:
,
  (2,2)
2 
 2
 11  5 1  3 
E :
,
  (3,1)
2 
 2
 1  11 7  1 
F :
,
  (6,4)
2 
 2
8
A(1, 7)
6
F (6, 4)
4
D(-2, 2)
C (11, 1)
2
x
−5
B(-5, -3)
5
−2
10
E (3, -1)
−4
DE  (2  3) 2  (2  1) 2  34 EF  (6  3) 2  (4  1) 2  34 DF  (6  2) 2  (4  2) 2  68  2 17
Since only two of the three sides are the same length, the triangle DEF is isosceles.
9. a) x = 65 (1/2 central angle), z = 115(cyclic quadrilateral), y = 65 (subtended by same arc as x)
b) x = 55 (180 in right triangle subtended by diameter), y = 70 (congruent triangles and angles)
c) x = 55 (1/2 central angle which equals arc AB), y = 125 (cyclic quad), z = 30 (E also 55 as
subtended by same arc as x, then 180 in triangle), arc ED= 60 (twice inscribed angle z),
arc AE = arc BD = 90 (arcs total 360)
d) arc BC= 80 (twice inscribed angles at A), arc AC = 156 (twice angle B), arc AB = 124 (360 total)
10. a) 2 x 2  2 y 2  12 x  4 y  12  0
(circle because same stretch on x and y)
x 2  6x  y 2  2 y  6
( x 2  6 x  9)  ( y 2  2 y  1)  6  9  1
( x  3) 2  ( y  1) 2  16
 x  3 
  y  1
 4    4   1
2
2
and
r=4
( x, y )  (4 x  3, 4 y  1)
Domain: [-1, 7]
b) 16 x 2  9 y 2  64 x  54 y  1  0
Range: [-5, 3]
(ellipse because different stretch on x and y)
16 x 2  64 x  9 y 2  54 y  1
16( x 2  4 x  4)  9( y 2  6 y  9)  1  64  81
Center (-2, 3)
16( x  2) 2  9( y  3) 2  144
( x, y )  (3x  2, 4 y  3)
( x  2) 2 ( y  3) 2

1
9
16
Domain: [-5, 1]
minor axis 6
Range: [-1, 7]
major axis 8
 x  2   y  3
 3    4  1

 

2
11.
Center (3,-1)
2
  y  1 
  x  5 
 3    2  1




2
 x  52
2
2
2
 2 ( y  1) 
 ( x  5) 
or it could have been 


  1 (rationaliz ed )

2
 3 


( y  1) 2
1
9
2
2
2( x  5) 2  9 y  1  18

2( x 2  10 x  25)  9( y 2  2 y  1)  18  0
2 x 2  9 y 2  20 x  18 y  41  0
 x  2
2
12. 
 2( y  3)  1

 7 
2
2
2
1

2

THINK RECIPROCAL S :  ( x  2)   ( y  3)  1
7

1

Geometric Proofs:


1.
Include the given information in your proofs
State the reason for each line of the proof
Statement
AD and BE bisect each other
AC = CD
BC = CE
BCA  ECD
BCA  ECD
AB  DE
Reason
given
definition bisect
definition bisect
vertically opposite angles
SAS
CPCTC
Statement
AB  AC
AD  AE
BAE  CAD
BAE  CAD
ABE  ACD
Reason
given
given
common angle
SAS
CPCTC
Statement
AE bisects BAC
Reason
given
given
definition of bisect
common side
SAS
CPCTC
straight line
straight line
supplements to equal angles
2.
3.
AB  AC
BAD  CAD
AD = AD
BAD  CAD
ADB  ADC
ADB supplementary to 1
ADC supplementary to 2
1  2
4.
Statement
1  2
FE  AC
CE = CE
AE  CF
EB  FD
BEA  DFC
B  D
Reason
given
given
common side
segment addition
given
SAS
CPCTC
5.
Statement
D is the midpoint of CE .
CD = DE
AC  AE, AB  AF
Reason
given
definition midpoint
given
BC = FE
C  E
BCD  FED
segment subtraction of congruent parts
given
SAS
CPCTC
BD  FD
6.
Statement
AB  BC
B  90
CD  BC
C  90
B  C
BA  CD
BC = BC
ABC  DCB
BCA  CBD
Reason
given
definition perpendicular
given
Statement
1  2
B  E
BC  DE
CD = CD
Reason
given
given
given
common side
segment addition
AAS
CPCTC
definition perpendicular
right angles
given
common side
SAS
CPCTC
7.
BD  EC
BDA  ECA
BDA  ECA
8.
Statement
AB  BE
B  90
EF  BE
E  90
B  E
BC  DE
CD = CD
BD  EC
AD  CF
ABD  FEC
A  F
Reason
given
definition perpendicular
given
definition perpendicular
right angles
given
common side
segment addition
given
SAS
CPCTC