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NSS Mathematics in Action 5A Full Solutions
98
5 Equations of Circles
Review Exercise 5 (p. 5.5)
1.
0
8
)7)(1(462
The equation x2 + 6x + 7 = 0 has two distinct realroots.
2.
0
)8)(18(4)24( 2
The equation 18x2 24x + 8 = 0 has one double realroot.
3. The graph of 362 xkxy has no x-intercepts.
i.e. The equation 0362 xkx has no real roots.
3
3612
0)3)((46
02
k
k
k
4.
082
033
yx
yx
)2(
)1(
(1) 2 + (2) :
2
0147
x
x
By substituting x = 2 into (1), we have
3
03)2(3
y
y
The solution of the simultaneous equations is(2, 3).
5.
xy
yx
4
04342 )2(
)1(
By substituting (2) into (1), we have
1or4
0)1)(4(
0432
yy
yy
yy
By substituting y = 4 into (1), we have
4
04)4(34
x
x
By substituting y = 1 into (1), we have
4
1
04)1(34
x
x
The solutions of the simultaneous equations are
(4, 4) and
1,4
1.
6. The equation of L is
i.e. 062
422
)4(2
11
yx
xy
xy
7. Slope of L
3
242
)1(3
The equation of L is
i.e. 0532
8233
)4(3
2)1(
yx
xy
xy
8. (a) Slope of L1
k
k
3
)(
3
Slope of L2
3
343
43
k
k
L1 L2
2
42
34
13
343
1ofslopeofSlope 21
k
k
kk
k
k
LL
(b)
(2)0332:
(1)02423:
2
1
yxL
yxL
(1) 3 + (2) 2:
6
07813
x
x
By substituting x = 6 into (1), we have
3
0242)6(3
y
y
The coordinates of the intersection are (6, 3).
Quick Practice
Quick Practice 5.1 (p. 5.7)(a) The equation of the circle is
i.e. 1
122
222
yx
yx
(b) The equation of the circle is
i.e. 5)3()2(
)5()]3([)2(22
222
yx
yx
Quick Practice 5.2 (p. 5.7)(a)
222
22
5)1()]5([
25)1()5(
yx
yx
Centre )1,5( , radius 5
(b)
222
22
22
)8()]1([)1(
8)]1([)1(
16)1(2)1(2
yx
yx
yx
Centre )1,1( , radius )22(or8
5 Equations of Circles
99
Quick Practice 5.3 (p. 5.8)Radius of the circle
50
)01()07( 22
OC
The equation of the circle is
50)1()7(
)50()1()7(22
222
yx
yx
Quick Practice 5.4 (p. 5.11)
(a) Centre
)2,4(
2
4,
2
8
Radius
5
5416
)5(2
4
2
822
(b)
02
14
2
5
018522
22
22
yxyx
yxyx
Centre
2,4
5
2
)4(,
225
Radius
4
916
81
2
14
16
25
2
1
2
4
22
52
2
Quick Practice 5.5 (p. 5.11)The equation represents an imaginary circle.
2
37
04
49
4
25
02
7
2
522
F
F
F
Quick Practice 5.6 (p. 5.12)
(a) The circle passes through P )4,3( .
By substituting )4,3( into
043222 cyxyx , we have
3
03
0)4(4)3(324)3( 22
c
c
c
(b) By substituting x = 0 and c = 3 into
043222 cyxyx , we have
3or1
0)3)(1(
034
034)0(3202
22
yy
yy
yy
yy
The coordinates of Q and R are (0, 1) and (0, 3)respectively.
(c)
2
4
32
4
2
32
(radii)
22
SRSQ
2
13
QR
SQ = SR = QRQRS is an equilateral triangle.
Quick Practice 5.7 (p. 5.14)Centre
)0,6(
2
44,
2
)9(3
ofpoint-mid
AB
Radius
5
25
4)3(
)]4(0[)]3(6[
22
22
The equation of the circle is (x + 6)2 + y2 = 25(or x2 + y2 + 12x + 11 = 0).
Alternative SolutionLet P(x, y) be a point on the circle apart from A and B.
AB is a diameter of the circle.AP BP ( in semi-circle)
01112
271216
)9)(3()4)(4(
1)9(
4
)3(
)4(
1ofslopeofSlope
22
22
xyx
xxy
xxyy
x
y
x
y
BPAP
Quick Practice 5.8 (p. 5.15)Let x2 + y2 + Dx + Ey + F = 0 be the equation of the circle.
The circle passes through (0, 0), (0, 8) and (4, 2).By substitution, we have
0
0)0()0(00 22
F
FED......(1)
648
0)8()0(80 22
FE
FED......(2)
2024
0)2()4(2)4( 22
FED
FED......(3)
NSS Mathematics in Action 5A Full Solutions
100
By substituting (1) into (2), we have
8
648
E
E
......(4)By substituting (1) and (4) into (3), we have
1
44
200)8(24
D
D
D
The equation of the circle is 0822 yxyx .
Quick Practice 5.9 (p. 5.15)(a) The circle touches the y-axis at P(0, 3).
The line joining the centre of the circle and P isperpendicular to the y-axis.
Let C(h, 3) be the centre of the circle.CQ = CP (radii)
2
5
52
4)21(
)35()1(
)35()1(
22
222
22
h
h
hhh
hh
hh
The equation of the circle is
4
25)3(
2
52
yx .
(b) From (a), coordinates of C
3,
2
5 and radius
2
5
4
97
44
81
)2(2
9
)31(2
52
22
22
CR
> radiusR(2, 1) lies outside the circle.
Quick Practice 5.10 (p. 5.17)Let (h, 0) be the coordinates of C.
CA = CB (radii)
4
328
3644444
36)2(4)2(
)60()2()]2(0[)]2([
22
22
2222
h
h
hhhh
hh
hh
Radius
40
436
)]2(0[)]2(4[ 22
CA
The equation of the circle is 40)4( 22 yx (or
).024822 xyx
Alternative Solution
Let 022 FEyDxyx be the equation of the circle.
The circle passes through two points A(2, 2) andB(2, 6).By substitution, we have
822
0)2()2()2()2( 22
EDF
FED
......(1)
4062
0)6()2(62 22
FED
FED
......(2)
The coordinates of the centre are
2
,2
EDand the
centre lies on the x-axis.
0
02
E
E
By substituting E = 0 and (1) into (2), we have
8
4084
40]8)0(22[)0(62
D
D
DD
By substituting E = 0 and D = 8 into (1), we have
24
8)0(2)8(2
F
The equation of the circle is 024822 xyx .
Quick Practice 5.11 (p. 5.24)
(a)
034
422 yxyx
y
)2(
)1(
By substituting (1) into (2), we have
0234
034442
22
xx
xx
For the equation 02342 xx ,
076)23)(1(442
02342 xx has no real roots.There are no intersections between the straight lineand the circle.
(b)
037127
1222 yxyx
xy
)4(
)3(
By substituting (3) into (4), we have
5or2
0)5)(2(
0107
050355
03712247144
037)12(127)12(
2
2
22
22
xx
xx
xx
xx
xxxxx
xxxx
By substituting x = 2 into (3), we have
3
1)2(2
y
By substituting x = 5 into (3), we have
9
1)5(2
y
The coordinates of the intersections between thestraight line and the circle are (2, 3) and (5, 9).
5 Equations of Circles
101
(c)
047128
013222 yxyx
yx
)6(
)5(
From (5), we have132 yx )7(
By substituting (7) into (6), we have
8
0)8(
06416
0320805
0471210416169524
04712)132(8)132(
2
2
2
22
22
y
y
yy
yy
yyyyy
yyyy
By substituting y = 8 into (7), we have
3
13)8(2
x
The coordinates of the intersection between the straightline and the circle are (3, 8).
Quick Practice 5.12 (p. 5.26)
(a)
076
622 yxyx
xy
)2(
)1(
By substituting (1) into (2), we have
037172
07663612
07)6(6)6(
2
22
22
xx
xxxxx
xxxx
For the equation 037172 2 xx ,
07)37)(2(4)17( 2 There are no intersections between the straight lineand the circle.
(b)
0125
8722 yxyx
yx
)4(
)3(
From (3), we have87 yx )5(
By substituting (5) into (4), we have
0132
0257550
01240356411249
012)87(5)87(
2
2
22
22
yy
yy
yyyyy
yyyy
For the equation 0132 2 yy ,
01)1)(2(4)3( 2 There are two intersections between the straight line
and the circle.
(c)
01184
053422 yxyx
yx
)7(
)6(
From (6), we have
4
53
yx )8(
By substituting (8) into (7), we have
012111025
017612880481625309
0176128)53(1616)53(
01184
534
4
53
2
22
22
22
yy
yyyyy
yyyy
yy
yy
For the equation 012111025 2 yy ,
0)121)(25(41102 There is only one intersection between the straight
line and the circle.
Quick Practice 5.13 (p. 5.28)
(a)
029212
122 yxyx
mxy
)2(
)1(
By substituting (1) into (2), we have
032)124()1(
029221212
029)1(212)1(
22
222
22
xmxm
mxxmxxmx
mxxmxx
......(*)
L1 is a tangent to C1.For the equation (*),
1or7
1
0)1)(17(
0167
08896
0)1(8)3(
0)32)(1(4)124(
0
2
22
22
22
mm
mm
mm
mmm
mm
mm
(b)
021204
0222 yxyx
cyx
)4(
)3(
From (3), we havecyx 2 )5(
By substituting (5) into (4), we have
0)214()124(5
021204844
02120)2(4)2(
22
222
22
ccycy
ycyyccyy
ycyycy
......(**)
L2 is a tangent to C2.For the equation (**),
3or47
0)3)(47(
014144
010520536244
0)214)(5()62(
0)214)(5(4)124(
0
2
22
22
22
cc
cc
cc
cccc
ccc
ccc
Quick Practice 5.14 (p. 5.29)
(a) Centre C
1),1(
2
2)(,
2
2
Radius
13
1111
)11(2
2
2
222
NSS Mathematics in Action 5A Full Solutions
102
(b) By substituting (2, 1) into ,0112222 yxyxwe have
L.H.S.
0
112414
11)1(2)2(2)1(2 22
R.H.S. = 0L.H.S. = R.H.S.A(2, 1) lies on S.
(c) Slope of CA
3
2
)1(2
11
Slope of the tangent =2
3
The equation of the tangent to S at A is
42
3
)2(2
3)1(
xy
xy
Quick Practice 5.15 (p. 5.31)
(a) (i)
042
122 kyxyx
xy
)2(
)1(
By substituting (1) into (2), we have
(*)0)5(82
044212
0)1(42)1(
2
22
22
kxx
kxxxxx
kxxxx
L1 is a tangent to S.For the equation (*),
3
248
040864
0)5)(2(48
02
k
k
k
k
(ii) From (a)(i), the equation of S is
034222 yxyx .
Radius
fig.)sig.3 to(cor.41.1
2
341
32
4
2
222
(b) Let (0, a) be the coordinates of A.A(0, a) lies on L1.By substituting (0, a) into the equation of L1, wehave
1
10
a
Coordinates of A = 1),0(
Coordinates of P
2),1(
2
4,
2
2
(c)
10
)12()01( 22
AP
In APB,ABP = 90 (tangent radius)
8
)2()10( 22
222
222
BPAPAB
BPABAP (Pyth. theorem)
fig.)sig.3 to(cor.83.2
8
AB
AC and AB are two tangents drawn to S from A.
fig.)sig.3 to(cor.83.2
8
)properties(tangent
ABAC
fig.)sig.3 to(cor.75.1
210
MPAPAM
Further Practice
Further Practice (p. 5.8)
1. (a) The equation of the circle is
i.e. 7
)7()0()0(22
222
yx
yx
(b) The equation of the circle is
i.e. 16)3(
4)]3([)0(22
222
yx
yx
(c) The equation of the circle is
i.e. 8)5()2(
)22()5()]2([22
222
yx
yx
(d) The equation of the circle is
i.e.4
9)5()4(
2
3)]5([)]4([
22
222
yx
yx
2. (a)222
22
6)3()]2([
36)3()2(
yx
yx
Centre )3,2( , radius 6
(b)222
22
5)1(
25)1(
yx
yx
Centre )1,0( , radius 5
(c)
2
22
22
22
3
16)0()0(
3
16
1633
yx
yx
yx
Centre )0,0( , radius
3
34or
3
16
5 Equations of Circles
103
(d)
222
22
22
)7()0()]2([
7)2(
284)42(
yx
yx
yx
Centre )0,2( , radius 7
3. Radius of C2
5
64362
1
)62()51(2
1
ofradius2
1
22
1
C
Centre of C2 = centre of C1 = (–1, –2)The equation of C2 is
25)2()1(
5)]2([)]1([22
222
yx
yx
Further Practice (p. 5.12)
1. (a) Centre
)3,2(
2
6,
2
)4(
Radius
5
)12(2
6
2
422
(b) Centre
)0,4(
2
0,
2
8
Radius
19
)3(2
0
2
822
(c) Centre
)6,0(
2
)12(,
2
0
Radius
4
202
12
2
022
(d)
0253
0820124422
22
yxyx
yxyx
Centre
2
5,
2
3
2
)5(,
2
3
Radius
2
26or
2
13
22
5
2
322
2. The equation represents a point circle.
5
041
02
4
2
222
k
k
k
3. (a) The circle passes through (0, 0).By substituting (0, 0) intox2 + y2 – 6x + py + q = 0, we have
0
0)0()0(600 22
q
qp
The circle passes through A (6, –5).By substituting (6, –5) and q = 0 intox2 + y2 – 6x + py + q = 0, we have
5
05362536
00)5()6(6)5(6 22
p
p
p
(b) The equation of S is x2 + y2 – 6x + 5y = 0.By substituting y = 0 intox2 + y2 – 6x + 5y = 0, we have
6or0
0)6(
06
0)0(5602
22
xx
xx
xx
xx
The coordinates of B are (6, 0).By substituting x = 0 intox2 + y2 – 6x + 5y = 0, we have
5or0
0)5(
05
05)0(602
22
yy
yy
yy
yy
The coordinates of C are (0, –5).
(c)
fig.)sig.3 to(cor.8.396
5tan
tan
OBC
OBC
OB
OCOBC
Further Practice (p. 5.17)
1. (a) Coordinates of M
)1,2(
2
20,
2
04
(b) 90AOBAB is a diameter of the circle.
(converse of in semi-circle)Radius
5
14
)10()24( 22
AM
Centre = (2, 1)The equation of the circle is(x – 2)2 + (y – 1)2 = 5(or x2 + y2 – 4x – 2y = 0).
NSS Mathematics in Action 5A Full Solutions
104
2. Let C(h, k) be the centre of the circle.The circle touches the x-axis at A and the y-axis atB(0, 5).CA x-axis and CB y-axisCoordinates of A = (h, 0) and k = 5
CA = CB (radii)
5
00
h
hk
Coordinates of C = (5, 5)Radius = CA
5
05
The equation of the circle is(x – 5)2 + (y – 5)2 = 25(or x2 + y2 – 10x – 10y + 25 = 0).
3. (a) Let C(0, k) be the centre of the circle.
2
5
208
44364416
)2()06()2()04(
(radii)
22
2222
k
k
kkkk
kk
CQCP
Radius
4
145
4
8116
2
52)04(
2
2
CP
The equation of the circle is
).0305or(
4
145
2
5
22
22
yyx
yx
(b) From (a), coordinates of
2
5,0C
Radius4
145
radius4
193
4
4936
12
5)60(
22
CR
R (6, 1) lies outside the circle.
Further Practice (p. 5.31)
1.
)2(0106
)1(922
cyxyx
xy
By substituting (1) into (2), we have
(*)0)9(22
0901068118
0)9(106)9(
2
22
22
cxx
cxxxxx
cxxxx
L intersects C.For the equation (*),
2
19
768
07284
0)9)(2(4)2(
02
c
c
c
c
2. (a)
)2(01810
)1(4222
yxyx
xy
By substituting (1) into (2), we have
3or1
0)3)(1(
032
015105
0132161016164
01)42(810)42(
2
2
22
22
xx
xx
xx
xx
xxxxx
xxxx
By substituting x = –1 into (1), we have
6
4)1(2
y
By substituting x = 3 into (1), we have
2
4)3(2
y
The coordinates of A and B are (–1, 6) and(3, –2) respectively.
(b) (i) Let C be the centre of the circle.
Coordinates of C
)4,5(
2
)8(,
2
)10(
Slope of CA3
1
)1(5
64
Slope of the tangent = 3The equation of the tangent to S at A is
93
)]1([36
xy
xy
(ii) Slope of CB
335
)2(4
Slope of the tangent3
1
The equation of the tangent to S at B is
13
1
)3(3
1)2(
xy
xy
3. (a)
)2(068184
)1(0422
yxyx
yx
From (1), we have)3(4 xy
5 Equations of Circles
105
By substituting (3) into (2), we have
2
0)2(
044
0686817
068)4(184)4(
2
2
2
22
x
x
xx
xx
xxxx
By substituting x = 2 into (3), we havey = 4(2) = 8Coordinates of P = )8,2(
)2(068184
)4(0131622
yxyx
yx
From (4), we have
)5(13
16xy
By substituting (5) into (2), we have
5
26
0)265(
067626025
0114924420425
06813
340
169
425
06813
16184
13
16
2
2
2
2
22
x
x
xx
xx
xx
xxxx
By substituting5
26x into (5), we have
5
32
5
26
13
16
y
Coordinates of Q
5
32,
5
26
(b) Slope of PQ
9
236
8
5
262
5
328
Slope of the perpendicular bisector PQ2
9
y-intercept of the perpendicular bisector = 0The equation of the perpendicular bisector of
PQ is xy2
9 .
Exercise
Exercise 5A (p. 5.18)Level 11. (a) The equation of the circle is
i.e. 16
422
222
yx
yx
(b) The equation of the circle is
i.e. 12)2(
)32()]2([22
222
yx
yx
(c) The equation of the circle is
i.e. 49)5()3(
7)5()3(22
222
yx
yx
(d) The equation of the circle is
i.e.9
16
3
1
3
2
3
4
3
1
3
2
22
222
yx
yx
2. (a)222
22
)24()]1([
24)1(
yx
yx
Centre )0,1( , radius )62(or24
(b)
2
22
22
22
3
5
3
25
2533
yx
yx
yx
Centre )0,0( , radius
3
35or
3
5
(c)222
22
10)8()]6([
100)8()6(
yx
yx
Centre )8,6( , radius 10
(d)
22
2
2
2
22
2
7
2
5)2(
4
49
2
5)2(
49)52()2(4
yx
yx
yx
Centre
2
5,2 , radius
2
7
3. (a) 02712822 yxyx
)6,4(
2
)12(,
2
)8(Centre
5
25
273616
272
12
2
8Radius
22
(b) 010922 yyx
2
9,0
2
9,
2
0Centre
NSS Mathematics in Action 5A Full Solutions
106
2
114
121
104
81
)10(2
9
2
0Radius
22
(c) 077222 yxyx
2
7,1
2
7,
2
)2(Centre
2
94
81
74
491
)7(2
7
2
2Radius
22
(d)
02
3132
0316422
22
22
yxyx
yxyx
2
3,1
2
3,
2
2Centre
2
35or
4
75
2
31
4
91
2
31
2
3
2
2Radius
22
4. (a) Radius of the circle
32
1616
)40()04( 22
CP
The equation of the circle is
32)4(
)32()]4([22
222
yx
yx
(b) Radius of the circle
2
10
4
9
4
1
2
512
2
322
CP
The equation of the circle is
2
5)1(
2
3
2
10)1(
2
3
22
2
22
yx
yx
5. Radius of the circle
27
7164
)7(2
8
2
422
Area of the circle
unitssq.27
unitssq.)27( 2
6. The equation represents an imaginary circle.
10
019
02
2
2
622
c
c
c
7. Let C be the centre of the circle.Coordinates of C (2, 3)Radius of the circle 6
(a)
radius2
154
225
364
81
)]3(3[2
5)2( 2
2
CA
3,2
5A lies outside the circle.
(b)
radius
20
416
)13()22( 22
CB
B(2, 1) lies inside the circle.
8. (a)
)3,1(
2
6,
2
)2(Centre
(b) Slope between A and the centre4
3
51
03
Slope between B and the centre4
3
)3(1
)6(3
A, B and the centre are colinear.AB is a diameter of the circle.
5 Equations of Circles
107
9. The equation represents a real circle.
4
25
04
94
02
3
2
422
k
k
k
k is positive.
4
250 k
We can suggest k 1 or 6 (or any positive value
satisfying4
250 k ).
10. Let C be the centre of the circle.
Coordinates of C
4,2
5
2
8,
2
)5(
2
9
)]4(4[2
52 2
2
CP
4
89where,
4
89
164
25
)(2
8
2
5Radius
22
cc
c
c
P(2, 4) lies outside the circle.
24
89
4
814
89
2
9
radius
c
c
c
CP
We can suggest c 3 or 4 (or any value of c less
than 2 and greater than4
89 ).
11. (a) By substituting A(1, 3) into x2 y2 6x ky 2 0,we have
6
183
023691
02)3()1(6)3(1 22
k
k
k
k
(b) From (a), the equation of S is x2 y2 6x 6y 2 0.
)3,3(
2
6,
2
6Centre
4
299
22
6
2
6Radius
22
12. (a)
5,2
5
2
)10(,
2
5Centre
(b)
c
c
c
4
125
254
25
2
10
2
5Radius
22
114
169
4
1252
13
4
125
c
c
c
13. (a) By substituting P(a, 2) into x2 y2 3x y 6 0,we have
3or0
0)3(
03
062)(322
22
aa
aa
aa
aa
(b) By substituting x 0 into x2 y2 3x y 6 0, wehave
2or3
0)2)(3(
06
06)0(302
22
yy
yy
yy
yy
The intersections of the circle and the y-axis are(0, 3) and (0, 2).
14. C is the mid-point of AB.
Coordinates of C
)3,0(
2
60,
2
22
13
94
)30()02(
Radius
22
CA
The equation of the circle is x2 (y 3)2 13(or x2 y2 6y 4 0).
15. C is the mid-point of AB.
Coordinates of C
3,2
3
2
)1(5,
2
)6(3
NSS Mathematics in Action 5A Full Solutions
108
4
97
44
81
)]5(3[32
3
Radius
22
CA
The equation of the circle is
4
97)3(
2
3 22
yx
(or x2 y2 3x 6y 13 0).
16. Let x2 y2 Dx Ey F 0 be the equation of the circle.The circle passes through O(0, 0), A(1, 0) andB(3, 2).By substitution, we have
0
0)0()0(00 22
F
FED……(1)
1
0)0()1(0)1( 22
FD
FED……(2)
1323
0)2()3(2)3( 22
FED
FED……(3)
By substituting (1) into (2), we have
1
1
D
D
By substituting D 1 and (1) into (3), we have
5
102
132)1(3
E
E
E
The equation of the circle is x2 y2 x 5y 0
2
13
2
5
2
1or
22
yx .
17. Let x2 y2 Dx Ey F 0 be the equation of the circle.The circle passes through P(0, 2), Q(0, 6) and R(2, 4).By substitution, we have
42
0)2()0(20 22
FE
FED……(1)
366
0)6()0(60 22
FE
FED……(2)
2042
0)4()2(42 22
FED
FED……(3)
(2) – (1) :
8
324
E
E
By substituting E 8 into (1), we have
12
4)8(2
F
F
By substituting E 8 and F 12 into (3), we have
0
02
2012)8(42
D
D
D
The equation of the circle is x2 y2 8y 12 0(or x2 (y 4)2 4).
18. Suppose the circle touches the x-axis at A. Then CA isperpendicular to the x-axis.Radius
2
3
2
30
CA
The equation of the circle is
)025310(or
4
9
2
3)5(
2
3
2
3)]5([
22
22
222
yxyx
yx
yx
19. (a) The circle touches the y-axis.Radius 5
The equation of the circle is (x 5)2 (y 4)2 25(or x2 y2 10x 8y 16 0).
(b) By substituting y 0 into x2 y2 10x 8y 16 0,we have
8or2
0)8)(2(
01610
016)0(81002
22
xx
xx
xx
xx
The coordinates of A and B are (2, 0) and (8, 0)respectively.
20. Let C be the centre of the circle.Coordinates of A (6, 0) and coordinates of B (0, 3)
OA OBAB is a diameter of the circle.
(converse of in semi-circle)The centre C of the circle is the mid-point of AB.
Coordinates of C
2
3,3
2
30,
2
06
4
45
4
99
02
3)63(
Radius
22
CA
The equation of the circle is
4
45
2
3)3(
4
45
2
3)3(
22
22
2
yx
yx
(or x2 y2 6x 3y 0)
5 Equations of Circles
109
Level 221. Let x2 y2 Dx Ey F 0 be the equation of the circle.
The circle passes through (0, 3), (4, 1) and (6, 1).By substitution, we have
93
0)3()0(30 22
FE
FED……(1)
174
0)1()4()1()4( 22
FED
FED
……(2)
376
0)1()6()1(6 22
FED
FED
……(3)(2) – (3) :
2
2010
D
D
By substituting D 2 into (2), we have
25
17)2(4
FE
FE
……(4)(1) – (4) :
4
164
E
E
By substituting E 4 into (4), we have
21
254
F
F
The equation of the circle isx2 y2 2x 4y 21 0 (or (x 1)2 (y 2)2 26).
22. Let x2 y2 Dx Ey F 0 be the equation of the circle.The circle passes through (1, 5), (6, 2) and(3, 11).By substitution, we have
265
0)5()1(51 22
FED
FED……(1)
4026
0)2()6()2()6( 22
FED
FED
……(2)
130113
0)11()3()11()3( 22
FED
FED
……(3)(1) – (2) :
2
1477
ED
ED
……(4)(3) – (2) :
303
9093
ED
ED
……(5)(4) – (5) :
8
324
E
E
By substituting E 8 into (4), we have
6
28
D
D
By substituting D 6 and E 8 into (1), we have
60
26)8(56
F
F
The equation of the circle isx2 y2 6x 8y 60 0 (or (x 3)2 (y 4)2 85).
23. (a)
26
251
)32()21(Radius 22
The equation of the circle is
26)2()1(
)26()]2([)1(22
222
yx
yx
(b) By substituting x 0 into (x 1)2 (y 2)2 26, wehave
7or3
0)7)(3(
0214
26441
26)2()10(
2
2
22
yy
yy
yy
yy
y
The coordinates of A and B are (0, 3) and (0, 7)respectively; or the coordinates of A and B are(0, 7) and (0, 3) respectively.
(c)
26
BCAC (radii)
10
)7(3
AB
By the cosine formula,
fig.)sig.3 to(cor.15713
1252
48
)26)(26(2
10)26()26(
))((2cos
222
222
ACB
BCAC
ABBCACACB
24. (a) Let C(h, 2h 2) be the centre of the circle.
1
2020
2520414444
)52()12()2(
)722()0()322()]2([
(radii)
2222
2222
2222
h
h
hhhhhhh
hhhh
hhhh
CBCA
Coordinates of C (1, 4)Radius
10
19
)34()]2(1[ 22
CA
The equation of the circle is(x 1)2 (y 4)2 10(or x2 y2 2x 8y 7 0).
(b)
radius
5
14
)45()13( 22
CP
P(3, 5) lies inside the circle.
25. (a) Let C(0, k) be the centre of the circle.
3
62
444121
)2()02()1()01(
(radii)
22
2222
k
k
kkkk
kk
CBCA
Coordinates of C (0, 3)
NSS Mathematics in Action 5A Full Solutions
110
Radius
5
41
)]3(1[)01( 22
CA
The equation of the circle is x2 (y 3)2 5(or x2 y2 6y 4 0).
(b)
radius
10
19
)]2(3[)]3(0[ 22
CQ
Q(3, 2) lies outside the circle.
26. (a) The circle touches the x-axis at A(–2, 0).The line joining the centre of the circle and A isperpendicular to the x-axis.
Let E(2, k) be the centre of the circle.
2
5
52
124
)]1([)]4(2[0
(radii)
22
22
k
k
kkk
kk
EBEA
The equation of the circle is
4
25
2
5)2(
2
5
2
5)]2([
22
222
yx
yx
(or x2 y2 4x 5y 4 0)(b) By substituting x 0 into
4
25
2
5)2(
22
yx , we have
4or12
3
2
5
4
9
2
5
4
25
2
52
2
22
yy
y
y
y
The coordinates of C and D are (0, 1) and(0, 4) respectively.
3
)4(1
CD
27. (a) Coordinates of C ),( rr
(b) The equation of the circle is (x r)2 (y r)2 r2.The circle passes through A(1, 8).By substituting (1, 8) into(x r)2 (y r)2 r2, we have
13or5
0)13)(5(
06518
166421
)8()1(
2
222
222
rr
rr
rr
rrrrr
rrr
The possible equations of the circle are
and
).01692626(or
169)13()13(
)0251010(or
25)5()5(
22
22
22
22
yxyx
yx
yxyx
yx
28. (a) Let (2k, k) be the coordinates of C.The circle touches the y-axis at P(0, –1).CP is perpendicular to the y-axis.k = –1
Coordinates of C
)1,2(
)1),1(2(
Radius = CP 2The equation of the circle is
4)1()2( 22 yx
(or x2 y2 4x 2y 1 0).
(b) By substituting y 0 into 4)1()2( 22 yx , wehave
32
3)2(
41)2(2
22
x
x
x
The x-intercepts of the circle are 32 and
32 .
29. (a) Let M be the mid-point of AB.
CM AB (line joining centre to mid-pt.of chord chord)
x-coordinate of C
52
82
Suppose the circle touches the y-axis at N.Then CN is perpendicular to the y-axis.
Radius
5 CN
(b) Let (5, k) be the coordinates of C.The equation of the circle is
(x 5)2 (y k)2 25 ……(1)By substituting (2, 0) into (1), we have
4
16
259
25)0()52(
2
2
22
k
k
k
k
For k 4,coordinates of C (5, 4)
radius523616
)]2(4[)15( 22
CP
But P(1, 2) lies inside the circle.k 4
For k 4,coordinates of C (5, 4)
radius20416
)]2(4[)15( 22
CP
P(1, 2) lies inside the circle.k 4
5 Equations of Circles
111
The equation of the circle is
25)4()5(
25)]4([)5(22
22
yx
yx
(or x2 y2 10x 8y 16 0)
30. (a) C1 cuts the x-axis at two points A and B.By substituting y 0 intox2 y2 8x 4y 15 0, we have
5or3
0)5)(3(
0158
015)0(4802
22
xx
xx
xx
xx
The coordinates of A and B are (3, 0) and (5, 0)respectively.
(b) Coordinates of S
)2,4(
2
4,
2
)8(
Let M and S′ be the mid-point of AB and the centre ofC2 respectively.SM AB and S′M AB
SS′ AB (line joining centre to mid-pt.of chord chord)
Let (4, k) be the coordinates of S′.C2 passes through A and S.
4
3
34
441
)]2([)44()0()34(22
2222
k
k
kkk
kk
Coordinates of S′
4
3,4
Radius of C2
4
516
91
04
3)34(
22
The equation of C2 is
03031622
16
25
16
9
2
3168
4
5
4
3)4(
22
22
222
yxyx
yyxx
yx
(c) Radius of
5
15416
152
4
2
822
1
C
The ratio of the area of C1 to that of C2
5:1616
25:5
4
5:)5(
22
31. (a) Centre of C1
)0,2(
2
0,
2
)4(
Radius of C1
1
34
32
0
2
422
Centre of C3 (2, 0)
Radius of
3
133
C
The equation of C3 is
054
944
3)2(
22
22
222
xyx
yxx
yx
(b) Centre of C2
)5,2(
2
10,
2
)4(
Radius of C2
2
25254
252
10
2
422
Distance between the centres of C2 and C3
5
)05()22( 22
Radius of C2 + radius of C3
5
32
Distance between the centres of C2 and C3
radius of C2 radius of C3
C2 and C3 touch each other.
32. (a) Slope of AB
2
126
24
Slope of BC
2106
)4(4
1)2(2
1ofslopeofSlope BCAB
AB BCABC is a right-angled triangle.
(b) AB BCAC is a diameter of the circle.(converse of in semi-circle)Centre of the circle
)1,6(
2
)4(2,
2
102
ofpoint-mid
AC
Radius of the circle
5
25
916
)21()26( 22
NSS Mathematics in Action 5A Full Solutions
112
The equation of the circumcircle of ABC is
25)1()6(
5)]1([)6(22
222
yx
yx
)012212or( 22 yxyx(c) Let (h, k) be the coordinates of D.
Slope of AD2
2
h
k
Slope of CD
10
410
)4(
h
kh
k
ABCD is a rectangle.Slope of AD slope of BC
and slope of CD slope of BA
i.e. 22
2
h
kand
2
1
10
4
h
k
422 hk and 1082 hk62 kh ……(1) and 182 kh ……(2)
(1) – (2) 2 :
6
305
k
k
By substituting k 6 into (1), we have
6
6)6(2
h
h
Coordinates of D )6,6(
33. (a) By substituting y 0 into x 3y 3 0, we have
3
03)0(3
x
x
Coordinates of A )0,3(
By substituting x 0 into 4x 3y 18 0, we have
6
0183)0(4
y
y
Coordinates of B )6,0(
)2.....(01834:
)1......(033:
2
1
yxL
yxL
(1) (2) :
3
0155
x
x
By substituting x 3 into (1), we have
2
0333
y
y
Coordinates of C )2,3(
(b) (i) Let x2 y2 Dx Ey F 0 be the equation ofthe circle.∵ The circle passes through A(3, 0), B(0, 6)
and C(3, 2).By substitution, we have
93
0)0()3(0)3( 22
FD
FED……(3)
366
0)6()0(60 22
FE
FED……(4)
1323
0)2()3(23 22
FED
FED……(5)
(4) – (3):
92
2763
ED
ED
……(6)(5) – (4): 2343 ED ……(7)
(6) 2 (7) :
1
55
D
D
By substituting D 1 into (6), we have
5
921
E
E
By substituting E 5 into (4), we haveF 6The equation of the circle isx2 y2 x 5y 6 0.
(ii) Centre
2
5,
2
1
2
)5(,
2
1
Radius
2
25or
2
5
2
25
64
25
4
1
)6(2
5
2
122
(c) Distance between the centre of the circle and thepoint D
circle theofradius2
52
25
4
49
4
1
62
5)1(
2
122
ABCD is a cyclic quadrilateral.
34. (a) The circle cuts the y-axis at two points Pand Q.By substituting x = 0 intox2 y2 4x 5y 4 0, we have
4or1
0)4)(1(
045
045)0(402
22
yy
yy
yy
yy
The coordinates of P and Q are (0, 1) and(0, 4) respectively.The circle touches the x-axis at the point R.By substituting y = 0 intox2 y2 4x 5y 4 0, we have
2
0)2(
044
04)0(540
2
2
22
x
x
xx
xx
The coordinates of R are (2, 0).
5 Equations of Circles
113
(b) Let (h, k) be the coordinates of S.
......(1)42
22
02
)4(0
)2(
0
ofslopeofSlope
khh
k
h
k
RQRS
PS RQ
......(2)222
112
1
0
)1(2
1ofslope
khh
kh
k
PS
(1) 2 (2) :
5
6
65
h
h
By substituting5
6h into (1), we have
5
8
45
62
k
k
The coordinates of S are
5
8,
5
6.
(c) (i) PS QSPQ is a diameter of the circle passingthrough P, Q and S.(converse of in semi-circle)
2
5,0
2
)4(1,0
ofpoint-midCentre PQ
2
32
)4(1Radius
The equation of the circle passing throughP, Q and S is
4
9
2
5
2
3
2
5)0(
22
222
yx
yx
(or 04522 yyx )
(ii) PS RSPR is a diameter of the circle passingthrough P, R and S.(converse of in semi-circle)
2
1,1
2
01,
2
)2(0
ofpoint-midCentre PR
2
5
142
1
)]1(0[)02(2
12
1Radius
22
PR
The equation of the circle passing throughP, R and S is
4
5
2
1)1(
2
5
2
1)]1([
22
222
yx
yx
(or 0222 yxyx )
35. (a) The circle cuts the x-axis at A and B.By substituting y = 0 into the equation of S, wehave
10or1
0)10)(1(
01011
010)0(71102
22
xx
xx
xx
xx
The coordinates of A and B are (1, 0) and(10, 0) respectively.The circle cuts the y-axis at P and Q.By substituting x = 0 into the equation of S, wehave
5or2
0)5)(2(
0107
0107)0(1102
22
yy
yy
yy
yy
The coordinates of P and Q are (0, 2) and (0, 5)respectively.
(b) (i) Let (h, 0) be the coordinates of R.Slope of PR slope of QB
42
12010
50
0
02
hh
h
The coordinates of R are (4, 0).Area of trapezium PRBQ
unitssq.21
unitssq.)2)(4(2
1)5)(10(
2
1
ofareaofarea
ORPOBQ
(ii) 2024 22 PR
125510 22 QBLet d be the distance between the parallel linesPR and QB.
NSS Mathematics in Action 5A Full Solutions
114
5
657
42
)5552(2
121
)12520(2
121
)(2
1 trapeziumofArea
d
d
d
dQBPRPRBQ
The required distance is
5
56or
5
6 .
(c) Let S and T be the mid-points of PQ and RBrespectively.
Coordinates of S
2
7,0
2
52,0
Coordinates of T
)0,7(
0,2
410
The equation of ST is
072
)7(2
1072
70
7
0
yx
xy
x
y
The required circle satisfies (1) and (2).Its centre lies on 072 yx and its radius
is5
3.
Consider 072 yx .
When x = 1,
3
0721
y
y
One possible equation of the required circle is
5
9)3()1( 22 yx .
When x = 3,
2
0723
y
y
One possible equation of the required circle is
5
9)2()3( 22 yx .
(or any other reasonable answers)
Exercise 5B (p. 5.32)Level 1
1. (a)
)2(01264
)1(22
yxyx
xy
By substituting (1) into (2), we have
2or3
0)2)(3(
06
01222
01264
2
2
22
xx
xx
xx
xx
xxxx
By substituting x = –3 into (1), we havey = –3
By substituting x = 2 into (1), we havey = 2The coordinates of the intersections between thestraight line and the circle are (–3, –3) and(2, 2).
(b)
)4(03246
)3(07222
yxyx
yx
From (3), we havey = –2x + 7 (5)
By substituting (5) into (4), we have
3
0)3(
096
045305
032288649284
032)72(46)72(
2
2
2
22
22
x
x
xx
xx
xxxxx
xxxx
By substituting x = 3 into (5), we have
1
7)3(2
y
The coordinates of the intersection between thestraight line and the circle are (3, 1).
(c)
)7(034123
)6(05422
yxyx
yx
From (6), we havey = 4x – 5 (8)
By substituting (8) into (7), we have
075
01198517
03460483254016
034)54(123)54(
2
2
22
22
xx
xx
xxxxx
xxxx
For the equation x2 – 5x + 7 = 0,
0
3
)7)(1(4)5( 2
There are no intersections between the straightline and the circle.
(d)
)10(0662
)9(0222
yxyx
yx
From (9), we havex = 2y (11)
By substituting (11) into (10), we have
06105
066)2(2)2(2
22
yy
yyyy
For the equation 5y2 – 10y + 6 = 0,
0
20
)6)(5(4)10( 2
There are no intersections between the straightline and the circle.
5 Equations of Circles
115
2. (a)
)2(0425
)1(03222
yxyx
yx
From (1), we havey = 2x – 3 (3)
By substituting (3) into (2), we have
07135
046459124
04)32(25)32(
2
22
22
xx
xxxxx
xxxx
For the equation 5x2 – 13x + 7 = 0,
0
29
)7)(5(4)13( 2
There are two intersections between the straightline and the circle.
(b)
)5(042522
)4(06322
yxyx
yx
From (4), we havex = –3y + 6 (6)
By substituting (6) into (5), we have
0465520
04230152727218
042)63(52)63(2
2
22
22
yy
yyyyy
yyyy
For the equation 20y2 – 55y + 46 = 0,
0
655
)46)(20(4)55( 2
There are no intersections between the straightline and the circle.
(c)
)8(0302
)7(022522
yxyx
yx
From (7), we have
)9(2
25
xy
By substituting (9) into (8), we have
0124429
01208204420254
0120)25(44)25(4
0302
252
2
25
2
22
22
22
xx
xxxxx
xxxx
xx
xx
For the equation 29x2 + 4x – 124 = 0,
0
40014
)124)(29(442
There are two intersections between the straightline and the circle.
(d)
)11(6
)10(0112422
yx
yx
From (10), we have
)12(2
114
xy
By substituting (12) into (11), we have
0978820
2412188164
62
114
2
22
22
xx
xxx
xx
For the equation 20x2 – 88x + 97 = 0,
0
16
)97)(20(4)88( 2
There are no intersections between the straightline and the circle.
3.
)2(044
)1(8222
cyxyx
xy
By substituting (1) into (2), we have
(*)0)32(205
0328464324
0)82(44)82(
2
22
22
cxx
cxxxxx
cxxxx
L does not intersect C.For the equation (*),
12
020640400
0)32)(5(420
02
c
c
c
4.
)2(0223
)1(322
yxyx
mxy
By substituting (1) into (2), we have
(*)05)34()1(
0262396
02)3(23)3(
22
222
22
xmxm
mxxmxxmx
mxxmxx
L is a tangent to C.For the equation (*),
2
11or
2
1
0)112)(12(
011244
011244
0202092416
0)5)(1(4)34(
0
2
2
22
22
mm
mm
mm
mm
mmm
mm
5. By substituting y = 0 into 4x2 + 4y2 – 12x – 8y + 9 = 0, wehave
(*)09124
09)0(812)0(442
22
xx
xx
For the equation (*),
0
)9)(4(4)12( 2
C touches the x-axis.From (*),
2
3
0)32(
091242
2
x
x
xx
The coordinates of the required point are
0,2
3.
NSS Mathematics in Action 5A Full Solutions
116
6.
)2(0644
)1(022
yxyx
cyx
From (1), we have)3(cxy
By substituting (3) into (2), we have
(*)0)64()82(2
064442
06)(44)(
22
222
22
ccxcx
cxxccxxx
cxxcxx
L intersects C at two points.For the equation (*),
4
0164
04832864324
0)64)(2(4)82(
0
2
2
22
22
c
c
cccc
ccc
We can suggest c = –1 or 1 (or any value of csatisfying c2 < 4).
7. Let S be the centre of the circle.
Coordinates of S
2
3,
2
3
2
)3(,
2
)3(
Slope of AS
7
3
52
3
32
3
Slope of the tangent3
7
The equation of the tangent is
3
44
3
7
)5(3
73
xy
xy
8. Let S be the centre of the circle.
Coordinates of S
2
1,0
2
)1(,
2
0
AS is a vertical line.The tangent is a horizontal line.The equation of the tangent is y = –2.
Alternative SolutionLet L: y = mx – 2 be the equation of the tangent.
)2(02
)1(222
yyx
mxy
By substituting (1) into (2), we have
(*)03)1(
02244
02)2()2(
22
222
22
mxxm
mxmxxmx
mxmxx
L is a tangent to C.For the equation (*),
0
0)0)(1(4)3(
022
m
mm
The equation of the tangent is y = –2.
9. Let S be the centre of the circle.
Coordinates of S
)1,2(
2
)2(,
2
)4(
AS is a horizontal line.The tangent is a vertical line.The equation of the tangent is x = 3.
10. Let L: y = mx + 2 be the tangent to the circle.
)2(038124
)1(222
yyyx
mxy
By substituting (1) into (2), we have
(*)018)48()1(
0382412444
038)2(124)2(
22
222
22
xmxm
mxxmxxmx
mxxmxx
L is a tangent to C.For the equation (*),
7or1
0)7)(1(
078
056648
07272166464
0)18)(1(4)]48([
0
2
2
22
22
mm
mm
mm
mm
mmm
mm
The equations of the tangents to the circle arey = x + 2 and y = 7x + 2.
11. Let L: y = –3x + c be the tangent to the circle.
)2(0524
)1(322
yxyx
cxy
By substituting (1) into (2), we have
(*)0)52()106(10
0526469
05)3(24)3(
22
222
22
ccxcx
cxxccxxx
cxxcxx
L is a tangent to C.For the equation (*),
15or5
0)15)(5(
07510
0300404
0200804010012036
0)52)(10(4)]106([
0
2
2
22
22
cc
cc
cc
cc
cccc
ccc
The equations of the tangents to the circle arey = –3x – 5 and y = –3x + 15.
5 Equations of Circles
117
12. (a) Slope of L2
2
)1(
2
L1 L2
Slope of L12
1 m
(b)
)2(07542:
)1(2
1:
22
1
yxyxC
cxyL
By substituting (1) into (2), we have
(*)0)754(4)4(45
0)754()4(4
5
0754224
1
0752
142
2
1
22
22
222
22
ccxcx
ccxcx
cxxccxxx
cxxcxx
L1 is a tangent to C.For the equation (*),
2
23or
2
17
0)232)(172(
0391124
0391124
0375205168
0)754(5)4(
0)]754(4)[5(4)]4(4[
0
2
2
22
22
22
cc
cc
cc
cc
cccc
ccc
ccc
13. (a) By substituting A(1, 1) into the equation of L1, wehave
1
021)1(
a
a
By substituting A(1, 1) into the equation of L2, wehave
17
010)1()1(7
b
b
(b) L1: x + y – 2 = 0 (1)L2: 7x – 17y + 10 = 0 (2)C: x2 + y2 – 12x + 28 = 0 (3)From (1), we have
y = –x + 2 (4)By substituting (4) into (3), we have
(*)0168
032162
0281244
02812)2(
2
2
22
22
xx
xx
xxxx
xxx
For the equation (*),
0
)16)(1(4)8( 2
L1 is a tangent to C.From (2), we have
)5(7
1017
yx
By substituting (5) into (3), we have
(**)
01156884169
023121768338
01372840142849100340289
01372)1017(8449)1017(
0287
101712
7
1017
2
2
22
22
22
yy
yy
yyyy
yyy
yy
y
For the equation (**),
0
)1156)(169(4)884( 2
L2 is a tangent to C.
Level 2
14. (a)
)2(0946
)1(03422
yxyx
pyx
From (1), we have
)3(3
4
pxy
By substituting (3) into (2), we have
(*)
0)8112()68(25
0811248548169
093
446
3
4
22
222
22
ppxpx
pxxppxxx
pxx
pxx
L is a tangent to C.For the equation (*),
28or8
0)28)(8(
022436
08064129636
081001200100369664
0)8112)(25(4)68(
0
2
2
22
22
pp
pp
pp
pp
pppp
ppp
(b) For p = 8,
5
7
0)75(
0497025
0]81)8(128[]6)8(8[25
2
2
22
x
x
xx
xx
By substituting5
7x and p = 8 into (3), we have
5
43
85
74
y
The coordinates of the intersection are
5
4,
5
7.
NSS Mathematics in Action 5A Full Solutions
118
For p = 28,
5
23
0)235(
052923025
0]81)28(1228[]6)28(8[25
2
2
22
x
x
xx
xx
By substituting5
23x and p = 28 into (3), we
have
5
16
3
285
234
y
The coordinates of the intersection are
5
16,
5
23.
15. (a) L1: 2x – y + q = 0 (1)L2: 2x + ry – 20 = 0 (2)C: 5x2 + 5y2 = 16 (3)From (1), we have
y = 2x + q (4)By substituting (4) into (3), we have
(*)0)165(2025
016520205
16)2(55
22
222
22
qqxx
qqxxx
qxx
L1 is a tangent to C.For the equation (*),
(rejected)4or4
16
1600100
01600500400
0)165)(25(4)20(
0
2
2
22
22
q
q
From (2), we have
)5(2
20
ryx
By substituting (5) into (3), we have
(**)01936200)205(
06420)40040(5
1652
205
22
222
22
ryyr
yryyr
yry
L2 is a tangent to C.For the equation (**),
(rejected)11or11
121
08801541280
0)1936)(205(4)200(
0
2
2
22
rr
r
r
rr
(b) For 042:1 yxL ,
from (*) in (a), we have
5
8
0)85(
0648025
016)4(5)4(2025
2
2
22
x
x
xx
xx
By substituting5
8x into the equation of L1, we
have
5
4
045
82
y
y
The coordinates of the intersection are
5
4,
5
8.
For 020112:2 yxL ,from (**) in (a), we have
25
44
0)4425(
019362200625
01936)11(200]20)11(5[
2
2
22
y
y
yy
yy
By substituting25
44y into the equation of L2, we
have
25
825
162
02025
44112
x
x
x
The coordinates of the intersection are
25
44,
25
8.
16. (a) Slope of L13
2
L1 L2
Slope of L22
3
A(0, 2) lies on L2.
The equation of L2 is 22
3 xy .
(b)
)2(22
3:
)1(01932:
2
1
xyL
yxL
By substituting (2) into (1), we have
2
132
13
01962
92
01922
332
x
x
xx
xx
By substituting x = 2 into (2), we have
5
2)2(2
3
y
The coordinates of the intersection of L1 and L2
are (2, 5).A(0, 2) is the centre of C.
L2 is the perpendicular to L1 passing through A.(2, 5) is the point of contact of L1 and C.
5 Equations of Circles
119
13
94
)52()20(Radius 22
The equation of C is
094
)13()2(22
222
yyx
yx
17. (a) C: x2 + y2 + 2x – 4y = 0 (1)C cuts the x-axis at O and A.By substituting y = 0 into (1), we have
2or0
0)2(
02
0)0(4202
22
xx
xx
xx
xx
The coordinates of A are (–2, 0).(b) Let S be the centre of C.
Coordinates of S
)2,1(
2
)4(,
2
2
Slope of OS
201
02
Slope of OB2
1
The equation of the tangent to C at O is2
xy .
Slope of SA
2
)2(1
02
Slope of AB2
1
The equation of the tangent to C at A is
1
2
)]2([2
1
xy
xy
(c)
)3(12
)2(2
xy
xy
By substituting (2) into (3), we have
1
122
x
xx
By substituting x = –1 into (2), we have
2
1y
The coordinates of B are
2
1,1 .
18. (a) The circle C cuts the y-axis at the two points Pand R.By substituting x = 0 intox2 + y2 – 6x – 10y + 21 = 0, we have
7or3
0)7)(3(
021102
yy
yy
yy
The coordinates of P and R are (0, 7) and (0, 3)respectively.
)2(021106:
)1(07:22
yxyxC
yxL
From (1), we havey = –x + 7 (3)
By substituting (3) into (2), we have
5or(rejected)0
0)5(
05
0102
021701064914
021)7(106)7(
2
2
22
22
xx
xx
xx
xx
xxxxx
xxxx
By substituting x = 5 into (3), we have
2
75
y
The coordinates of Q are (5, 2).
(b) Area of PQRunitssq.10
unitssq.)05)(37(2
1
19. (a) L passes through the point P(2, 0).By substituting P(2, 0) into the equation of L,we have
2
002
b
b
S passes through the point P(2, 0).By substituting P(2, 0) into the equation of S, wehave
20
0)0(4)2(802 22
c
c
(b)
)2(02048:
)1(02:22
yxyxC
yxL
From (1), we havey = –x + 2 (3)
By substituting (1) into (2), we have
(rejected)2or6
0)2)(6(
0124
02482
02084844
020)2(48)2(
2
2
22
22
xx
xx
xx
xx
xxxxx
xxxx
By substituting x = –6 into (3), we have
8
2)6(
y
Coordinates of Q )8,6(
PQ
)28(or128
6464
)08()26( 22
NSS Mathematics in Action 5A Full Solutions
120
(c) Let M be the mid-point of PQ.
Coordinates of M
4),2(
2
80,
2
)6(2
Coordinates of centre C
2),4(
2
4)(,
2
8
M is the mid-point of PQ.CM PQ (line joining centre to mid-pt. of
chord chord)Distance between the centre C and PQ
)22or(8
)24()]4(2[ 22
20. (a) (i) By substituting y = x into x2 + y2 – ky + 2 = 0,we have
(*)022
022
22
kxx
kxxx
L1 is a tangent to the circle S.For the equation (*),
(rejected)4or4
16
0)2)(2(4)(
0
2
2
kk
k
k
(ii) S: x2 + y2 – 4y + 2 = 0
Radius of S
2
24
22
4
2
022
(b) Since L2 is a tangent to the circle S,CA CP (tangent radius)
i.e. OCA = 90Slope of L1 = 1
COA
54
1tan90
ofninclinatio901
1L
CAO
45
4590180
)ofsum(180 COAOCA
COA is a right-angled isosceles triangle.
(c) Coordinates of P
2),0(
2
)4(,
2
0
2OP
22
PCOPOC
22
s)equalopp.sides(
)propertiestangent(
OC
ACAB
21. (a)
)2(010464:
)1(02632:22
yxyxC
yxL
From (1), we have
)3(132
3 yx
By substituting (3) into (2), we have
(*)03612
0117394
13
01046526
169394
9
01046132
3413
2
3
2
2
22
22
yy
yy
yy
yyy
yyyy
For the equation (*),
0
)36)(1(4122
L is a tangent to S.
From (*), we have
6
0)6( 2
y
y
By substituting y = 6 into (3), we have
4
13)6(2
3
x
Coordinates of P )6,4(
(b) Let (x, y) be the coordinates of Q.
Coordinates of C
)3,2(
2
6,
2
)4(
By the mid-point formula, we have
12and82
63and
2
)4(2
yx
yx
Coordinates of Q )12,8(
(c) L1 // L and L1 is a tangent to S.L1 must pass through Q.
3
2
)3(
2ofSlope
L
3
2
ofslopeofSlope 1
LL
The equation of L1 is
3
52
3
2
)8(3
2)12(
xy
xy
22. (a) C passes through the point A(–4, 0).By substituting A(–4, 0) into the equation of C,we have
44
164
0)0()4(0)4( 22
FD
FD
FED
5 Equations of Circles
121
C passes through the point B(0, 4).By substituting B(0, 4) into the equation of C,we have
44
164
0)4()0(40 22
FE
FE
FED
(b)
)2(044
44
:
)1(083:
22
FyF
xF
yxC
yxL
From (1), we havex = –3y – 8 (3)
By substituting (3) into (2), we have
(*)0)32()32(10
044
322
124
364489
044
)83(44
)83(
2
22
22
FyFy
FyyF
F
yyF
yyy
FyF
yF
yy
C touches the straight line L.For the equation (*),
8or32
0)8)(32(
0)32)(10(4)32(
02
FF
FF
FF
For F = 32,
1244
32
1244
32
E
D
Centre 6),6(2
)12(,
2
12
Radius
40
323636
322
12
2
1222
Distance from O to the centre
radius
72
)06()06( 22
O lies inside C.F 32
For F = –8,
244
)8(
244
8
E
D
Centre 1),1(2
)2(,
2
2
Radius
10
)8(2
2
2
222
Distance from O to the centre
radius
2
)01()01( 22
O lies inside C.F = –8The equation of C is x2 + y2 + 2x – 2y – 8 = 0.
Revision Exercise 5 (p. 5.37)Level 11. (a) The equation of the circle is
i.e. 36
622
222
yx
yx
(b) The equation of the circle is
i.e. 32)2()5(
)24()2()]5([22
222
yx
yx
2. (a)222
22
5
025
yx
yx
Centre )0,0(
Radius 5
(b) Centre
)4,3(
2
)8(,
2
6
Radius
5
25
169
02
8
2
622
(c)
02
32
2
3
034322
22
22
yxyx
yxyx
Centre
1,4
3
2
)2(,
2
2
3
Radius
4
716
49
2
31
16
9
2
3
2
2
22
32
2
NSS Mathematics in Action 5A Full Solutions
122
(d)
03
4
3
4
3
8
044833
22
22
yxyx
yxyx
Centre
3
2,
3
4
23
4
,2
3
8
Radius
3
22or
9
8
3
4
9
4
9
16
3
4
23
4
23
822
3. (a) Radius
4
3
164
259
162
5
2
622
The equation represents an imaginary circle.
(b)
02
419
04121822
22
22
yxyx
yxyx
Radius
02
41
4
1
4
81
2
41
2
1
2
922
The equation represents a point circle.
4. Let S be the centre of C.
2
1,
2
5
2
)1(,
2
5S
Radius
2
25
64
1
4
25
)6(2
1
2
522
(a) SP
radius2
13
02
10
2
522
P(0, 0) lies inside the circle.
(b) SQ
radius2
25
4
49
4
1
42
1)2(
2
522
Q(–2, 4) lies on the circle.
(c) SR
radius2
29
4
9
4
49
22
11
2
522
R(1, 2) lies outside the circle.
5. (a)
)2(0591210
)1(0322
yxyx
yx
From (1), we have)3(3 xy
By substituting (3) into (2), we have
4
0)4(
0168
032162
05936121096
059)3(1210)3(
2
2
2
22
22
x
x
xx
xx
xxxxx
xxxx
By substituting x = 4 into (3), we have
7
34
y
The coordinates of the intersection between thestraight line and the circle are (4, 7).
(b)
)5(015124
)4(05322
yxyx
yx
From (4), we have)6(53 yx
By substituting (6) into (5), we have
2or1
0)2)(1(
023
0203010
01512201225309
01512)53(4)53(
2
2
22
22
yy
yy
yy
yy
yyyyy
yyyy
5 Equations of Circles
123
By substituting y = –1 into (6), we have
2
5)1(3
x
By substituting y = –2 into (6), we have
1
5)2(3
x
The coordinates of the intersections between thestraight line and the circle are (–2, –1) and(1, –2).
6. (a)
)2(016
)1(042322
xyx
yx
From (1), we have
)3(22
3 xy
By substituting (3) into (2), we have
(*)034
13
016464
9
01622
3
2
22
22
x
xxxx
xxx
For the equation (*),
039
)3(4
13402
There are no intersections between the straightline and the circle.
(b)
)5(019233
)4(085222
yyx
yx
From (4), we have
)6(42
5 yx
By substituting (6) into (5), we have
(**)0483
029584
87
0192348604
75
0192342
53
2
2
22
22
yy
yy
yyyy
yyy
For the equation (**),
016
)4)(3(4)8( 2
There are two intersections between the straightline and the circle.
7.
)2(022
)1(1222
cyxyx
xy
By substituting (1) into (2), we have
(*)0)3(65
0242144
0)12(22)12(
2
22
22
cxx
cxxxxx
cxxxx
L intersects S at two points.For the equation (*),
5
6
02024
0)3)(5(46
02
c
c
c
8. (a) Coordinates of C = (1, 3)Radius = 3
The equation of the circle is(x – 1)2 + (y – 3)2 = 9(or x2 + y2 – 2x – 6y + 1 = 0).
(b) Coordinates of C = (–3, 2)Radius = –3 – (–7) = 4
The equation of the circle is
16)2()3(
4)2()]3([(22
222
yx
yx
)0346or( 22 yxyx
9. Radius
2
5
12
3)11(
22
PQ
The equation of the circle is
4
25
2
3)1(
2
5
2
3)1(
22
222
yx
yx
10. AB is a diameter of the circle.Centre
)3,1(
2
15,
2
)6(4
ofpoint-mid
AB
Radius
29
1162
1
)15()]6(4[2
12
1
22
AB
The equation of the circle is(x + 1)2 + (y – 3)2 = 29(or x2 + y2 + 2x – 6y – 19 = 0).
11. (a) Let x2 + y2 + Dx + Ey + F = 0 be the equation of thecircle.
The circle passes through the points A (1, 0),B (12, 0) and C (0, 2).By substitution, we have
1
0)0()1(01 22
FD
FED
……(1)
14412
0)0()12(012 22
FD
FED
……(2)
NSS Mathematics in Action 5A Full Solutions
124
42
0)2()0(20 22
FE
FED
……(3)(2) – (1) :
13
14311
D
D
By substituting D = –13 into (1), we have
12
113
F
F
By substituting F = 12 into (3), we have
8
4122
E
E
The equation of the circle is
4
185)4(
2
13or
012813
22
22
yx
yxyx
.
(b) By substituting x = 0 into x2 + y2 – 13x – 8y +12 = 0,we have
6or(rejected)2
0)6)(2(
0128
0128)0(1302
22
yy
yy
yy
yy
The coordinates of the required intersection are(0, 6).
12. (a) Let x2 + y2 + Dx + Ey + F = 0 be the equation of thecircle.
The circle passes through (–2, 2), (3, 2) and(3, –4).By substitution, we have
822
0)2()2(2)2( 22
FED
FED
……(1)
1323
0)2()3(23 22
FED
FED
……(2)
2543
0)4()3()4(3 22
FED
FED
……(3)(2) – (1) :
1
55
D
D
(2) – (3) :
2
126
E
E
By substituting D = –1 and E = 2 into (1), wehave
14
8)2(2)1(2
F
F
The equation of the circle is
4
61)1(
2
1or
0142
22
22
yx
yxyx
(b) Centre
1,2
1
2
2,
2
)1(
Radius
2
61or
4
61
1414
1
)14(2
2
2
122
13. Distance between (2, –5) and (–1, –1)
5
25
169
)]1(5[)]1(2[ 22
The equation of the circle is(x – 2)2 + (y + 5)2 = 26 or (x – 2)2 + (y + 5)2 = 27(or any equation in the form (x – 2)2 + (y + 5)2 = r2,where r > 5).
14. Let x2 + y2 + Dx + Ey + F = 0 be the equation of the circle.The circle passes through A(2, 2) and B(–2, –2).By substitution, we have
822
0)2()2(22 22
FED
FED
……(1)
822
0)2()2()2()2( 22
FED
FED
……(2)(1) + (2) :
8
162
F
F
By substituting F = –8 into (1), we have
DE
ED
8822
The equation of the circle isx2 + y2 = 8 or x2 + y2 + x – y – 8 = 0(or any equation in the formx2 + y2 + kx – ky – 8 = 0, where k is a realnumber).
15. (a) Let r be the radius of the circle.The equation of the circle is x2 + (y – k)2 = r2 ……(*)
The circle passes through (6, 2) and (–1, –5).By substitution, we have
1
1414
2610404
2610and404
)5()1(and)2(6
22
2222
222222
k
k
kkkk
rkkrkk
rkrk
(b) By substituting k = 1 into (*), we havex2 + (y – 1)2 = r2
The circle passes through (6, 2).
37
)12(62
222
r
r
The equation of C is x2 + (y – 1)2 = 37(or x2 + y2 – 2y – 36 = 0).
16. Let (h, h) be the centre of the circle and r be the radius ofthe circle.The equation of the circle is
)1()()( 222 rhyhx
5 Equations of Circles
125
The circle passes through (3, 0) and (0, –2).By substitution, we have
)2(269
)0()3(22
222
rhh
rhh
)3(244
)2()0(22
222
rhh
rhh
(3) – (2) :
2
1
0105
h
h
By substituting2
1h into (2), we have
2
13
2
12
2
169
2
22
r
r
The equation of the circle is
).06or(
2
13
2
1
2
1
22
22
yxyx
yx
17. (a) By substituting y = 0 into x2 + y2 – 2x – 4y + 1 = 0,we have
(*)012
01)0(4202
22
xx
xx
For the equation (*),
0
)1)(1(4)2( 2
The circle C touches the x-axis.
(b) Centre of C
)2,1(
2
)4(,
2
)2(
Radius
2
12
4
2
222
Distance between A(–5, 0) and the centre of C
radius
40
)20()15( 22
A(–5, 0) lies outside C.
18. Let S be the centre of the circle.Coordinates of S = (–2, 0)
Slope of AS
1
)2(3
05
Slope of the tangent = 1The equation of the tangent to C at A is
8
3)5(
xy
xy
19. Let S be the centre of the circle.
Coordinates of S
2,2
5
2
)4(,
2
5
Slope of AS
4
25
1
24
Slope of the tangent4
1
The equation of the tangent to C at A is
4
15
4
1
)]1([4
1)4(
xy
xy
20. Let y = mx + 4 be the equation of the tangent.
)2(042
)1(422
yyx
mxy
By substituting (1) into (2), we have
(*)02010)1(
0482168
04)4(2)4(
22
222
22
mxxm
mxmxxmx
mxmxx
y = mx + 4 is the equation of the tangent.For the equation (*),
2
4
8020
08080100
0)20)(1(4)10(
0
2
2
22
22
m
m
m
mm
mm
The equations of the tangents are y = 2x + 4 andy = –2x + 4.
21. (a) C cuts the x-axis at the points A and B.By substituting y = 0 intox2 + y2 – 5x + ky + 4 = 0, we have
4or1
0)4)(1(
045
04)0(502
22
xx
xx
xx
kxx
The coordinates of A and B are (1, 0) and (4, 0)respectively or the coordinates of A and B are(4, 0) and (1, 0) respectively.
(b) By substituting x = 0 intox2 + y2 – 5x + ky + 4 = 0, we have
(*)04
04)0(502
22
kyy
kyy
C touches the y-axis at the point T.For the equation (*),
4
0)4)(1(4
02
k
k
NSS Mathematics in Action 5A Full Solutions
126
For k = –4,
2
0)2(
0442
2
y
y
yy (from (*))
The corresponding coordinates of T are (0, 2).For k = 4,
2
0)2(
0442
2
y
y
yy (from (*))
The corresponding coordinates of T are (0, –2).
22. (a)
)2(014
)1(03222
yxyx
ayx
From (1), we have
)3(22
3
ayx
By substituting (3) into (2), we have
(*)0124
72
3
4
13
012642
3
4
9
0122
34
22
3
22
22
2
22
aa
ya
y
yayya
ya
y
ya
yya
y
L is a tangent to C.For the equation (*),
9or4
0)9)(4(
0365
0365
013264
134921
4
9
01244
1347
2
3
0
2
2
22
22
aa
aa
aa
aa
aaaa
aaa
(b) For a = –4,
2
0)2(
044
013134
13
01)4(24
)4(7
2
)4(3
4
13
2
2
2
22
y
y
yy
yy
yy (from (*))
By substituting y = 2 and a = –4 into (3), we have
12
)4()2(
2
3
x
The coordinates of the intersection are (–1, 2).
For a = 9,
1
0)1(
012
04
13
2
13
4
13
01)9(24
97
2
)9(3
4
13
2
2
2
22
y
y
yy
yy
yy (from (*))
By substituting y = –1 and a = 9 into (3), we have
32
9)1(
2
3
x
The coordinates of the intersection are (–3, –1).
23. (a) Both 01:1 pyxL and 012
:2 yq
xL are in
the form 01 kyx .
)2(024
)1(0122
yxyx
kyx
From (1), we have)3(1 kyx
By substituting (3) into (2), we have
(*)05)26()1(
024412
02)1(4)1(
22
222
22
ykyk
ykyykyyk
ykyyky
L1 and L2 are tangents to C.For the equation (*),
2or2
1
0)2)(12(
0232
0162416
0202042436
0)5)(1(4)26(
0
2
2
22
22
kk
kk
kk
kk
kkk
kk
0and0
02
and01
0ofslopeand0ofSlope 21
qp
qp
LL
12
1
2and2
q
qp
(b) For L1, by substituting k = –2 into (*), we have
1
0)1(
012
05105
05]2)2(6[]1)2[(
2
2
2
22
y
y
yy
yy
yy
By substituting y = 1 and k = –2 into (3), we have
1
1)1)(2(
x
The coordinates of the intersection are (1, 1).
5 Equations of Circles
127
For L2, by substituting2
1k into (*), we have
2
0)2(
044
0554
5
0522
161
2
1
2
2
2
22
y
y
yy
yy
yy
By substituting y 2 and2
1k into (3), we have
0
1)2(2
1
x
The coordinates of the intersection are (0, 2).
24. Slope of4
31 L
L L1
33
443
4ofSlope
bb
L
)2(......0828:
)1......(034:22 yxyxC
cyxL
From (1), we have
33
4 cxy ……(3)
By substituting (3) into (2), we have
......(*)083
2
93
32
9
8
9
25
083
2
3
88
99
8
9
16
0833
428
33
4
22
222
22
ccxcx
cxx
ccxxx
cxx
cxx
L is a tangent to C.For the equation (*),
2or28
0)2)(28(
05626
09
224
9
104
9
4
09
800
27
200
81
100
9
1024
27
512
81
64
083
2
99
254
3
32
9
8
0
2
2
22
22
cc
cc
cc
cc
cccc
ccc
Level 2
25. (a) Radius2
7
Centre
2
7,
2
7
The equation of the circle is222
2
7
2
7
2
7
yx
4
49
2
7
2
722
yx
04
4977or 22 yxyx
(b) Let (0, k) be the centre of the circle.Radius kThe equation of the circle is
222 )( kkyx ……(1)
Slope of the tangent with inclination 60
3
60tan
The equation of the tangent is
33
)]3([3
xy
xy
……(2)
By substituting (2) into (1), we have
0)69()3(324
)3(3)3(23
)33(
2
2222
222
kxkx
kkxkxx
kkxx
……(*)(2) is the tangent of the circle.For the equation (*),
1or(rejected)3
0)1)(3(
032
081269
096144)69(12
0)69)(4(4)]3(32[
0
2
2
2
2
kk
kk
kk
kkk
kkk
kk
The equation of the circle is
)02(or1)1( 2222 yyxyx .
26. (a) Slope of AB
124
02
Slope of the perpendicular bisector of AB 1
Mid-point of AB
)1,3(
2
20,
2
42
The equation of the perpendicular bisector of ABis
4
)3(1
xy
xy
(b) The circle passes through A and B.The centre lies on the perpendicular bisector ofAB.The centre is the intersection of theperpendicular bisector of AB and L : y 2x 11.
NSS Mathematics in Action 5A Full Solutions
128
)2......(112
)1......(4
xy
xy
By substituting (1) into (2), we have
5
153
1124
x
x
xx
By substituting x 5 into (1), we have
1
45
y
Centre (5, 1)
Radius
10
)01()25( 22
The equation of the circle is
).016210(or
10)1()5(22
22
yxyx
yx
27. Let C(h, 7 h) be the centre of the circle.The circle touches the y-axis.Radius h
The equation of the circle is222 )]7([)( hhyhx ……(1)
By substituting A(1, 1) into (1), we have
13or5
0)13)(5(
06518
641621
)8()1(
)71()1(
2
222
222
222
hh
hh
hh
hhhhh
hhh
hhh
The equations of the circles are
25)2()5(
5)]57([)5(22
222
yx
yx
or
169)6()13(
13)]137([)13(22
222
yx
yx
28. (a) Slope of4
1L
AB L (tangent radius)Slope of AB 4The equation of the straight line is
74
2
341
xy
xy
(b) B is the intersection of 0114: yxL and
AB : y 4x 7.
)2......(74
)1......(0114
xy
yx
By substituting (2) into (1), we have
1
1717
011)74(4
x
x
xx
By substituting x 1 into (2), we have
3
7)1(4
y
Coordinates of B (1, 3)
4
17
)31()1(2
3
Radius
22
AB
The equation of C is
)0123(or
4
17)1(
2
3
4
17)1(
2
3
22
22
22
yxyx
yx
yx
29. (a) Slope of AC
134
23
Slope of BC
147
30
Slope of AC slope of BC 1AC is perpendicular to BC.
(b) AC BCAB is a diameter of the circle. (converse of in
semi-circle)Centre
)1,5(
2
02,
2
73
ofpoint-mid
AB
Radius
5
)02()73(2
12
1
22
AB
The equation of the circle is
).021210(or
5)1()5(22
22
yxyx
yx
(c) The circle cuts the x-axis at two points B and D.By substituting y 0 into
02121022 yxyx , we have
(rejected)7or3
0)7)(3(
02110
021)0(21002
22
xx
xx
xx
xx
Coordinates of D )0,3(
30. (a)
)2......(
)1......(222 ayx
cmxy
By substituting (1) into (2), we have
......(*)0)(2)1(
02
)(
2222
22222
222
acmcxxm
acmcxxmx
acmxx
5 Equations of Circles
129
L is a tangent to C.For the equation (*),
......(**))1(
0)1(4444
0))(1(4)2(
0
222
2222222
2222
mac
amcmccm
acmmc
(b) (i) By substituting a2 90 and m 3 into (**), wehave
30
900
)31(902
22
c
c
c
The equations of the tangents arey 3x 30 and y 3x 30.
(ii) By substituting a2 90 and c 10 into (**), wehave
3
19
1
1090
)1(9010
2
2
22
m
m
m
m
The equations of the tangent are
103
xy and 10
3
xy .
31. (a) Slope of L3
4
AB L (tangent radius)
Slope of AB4
3
Centre of C
2
1,1
2
1,
2
2
The equation of AB is
4
1
4
34
3
4
3
2
1
)]1([4
3
2
1
xy
xy
xy
(b)
)2......(4
1
4
3:
)1......(0234:
xyAB
cyxL
By substituting (2) into (1), we have
25
454
5
4
25
024
3
4
25
024
1
4
334
cx
cx
cx
cxx
By substituting25
45 cx
into (2), we have
25
310100
12404
1
100
1215
4
1
25
45
4
3
c
c
c
cy
Coordinates of A
25
310,
25
45 cc
(c) By substituting A
25
310,
25
45 cc into
02: 22 cyxyxC , we have
025
310
25
452
25
310
25
4522
c
cccc
5
0)5(
02510
062525025
062575250200250
960100164025
2
2
2
22
c
c
cc
cc
ccc
cccc
(d)
)4......(052
)3......(4
1
4
3
22 yxyx
xy
By substituting (3) into (4), we have
(rejected)1or3
0)1)(3(
032
016
75
8
25
16
25
054
1
4
32
16
1
8
3
16
9
054
1
4
32
4
1
4
3
2
2
22
22
xx
xx
xx
xx
xxxxx
xxxx
By substituting x 3 into (3), we have
24
1)3(
4
3
y
Coordinates of B (3, 2)The equation of the tangent to C at B is
63
4
)]3([3
4)2(
xy
xy
32. (a) Slope of L tan 45 1The equation of L is 1 xy .
(b) Coordinates of C
)3,4(
2
)6(,
2
)8(
BC L (tangent radius)Slope of BC 1
NSS Mathematics in Action 5A Full Solutions
130
The equation of BC is
7
)4(3
xy
xy
)2......(7
)1......(1
xy
xy
By substituting (1) into (2), we have
3
71
x
xx
By substituting x 3 into (1), we havey 3 1 4Coordinates of B (3, 4)S passes through B(3, 4).By substituting B(3, 4) into
06822 kyxyx , we have
23
0)4(6)3(843 22
k
k
(c) Radius of S 2232
6
2
822
The centre of S1 lies on the straight line passingthrough B and C.Let (h, h 7) be the centre of S1.
The equation of S1 is
2)7()( 22 hyhx ……(3)By substituting B(3, 4) into (3), we have
(rejected)4or2
0)4)(2(
086
29669
2)74()3(
2
22
22
hh
hh
hh
hhhh
hh
The equation of S1 is
2)5()2(
2)72()2(22
22
yx
yx
)027104(or 22 yxyx
33. (a) The circle cuts the x-axis at A and B.By substituting y 0 into
086222 yxyx , we have
2or4
0)4)(2(
082
08)0(6202
22
xx
xx
xx
xx
Coordinates of A )0,4(
Coordinates of B )0,2(
Coordinates of C
)3,1(
2
)6(,
2
2
(b) Slope of AC 1)4(1
03
Slope of the tangent to S at D 1Let (h, k) be the coordinates of D.
C is the mid-point of AD.
6and22
03and
2
41
kh
kh
Coordinates of D (2, 6)The equation of the tangent to S at D is
8
)2(6
xy
xy
(c) (i)
45
1tan
(ii) The equation of line passing through C and thecentre of S1 is
2
)]1([3
xy
xy
The centre of S1 lies on the x-axis.
By substituting y 0 into 2 xy , we
have x 2Centre of S1 (2, 0)
8
)03()21(Radius 22
The equation of S1 is
).0144(or
18)2(22
22
xyx
yx
(iii)
......(2)18)2(
......(1)822 yx
xy
By substituting (1) into (2), we have
......(*)02510
050202
18641644
18)8()2(
2
2
22
22
xx
xx
xxxx
xx
For the equation (*),
0
)25)(1(4)10( 2
The tangent to S at D is also the tangent toS1.
34. (a) The equation of L is
mmxy
xmy
71
)7(1
(b) (i)
......(2)20:
......(1)71:22 yxC
mmxyL
By substituting (1) into (2), we have
0)191449()71(2)1(
20)71()71(2
20)71(
222
2222
22
mmxmmxm
mxmmxmx
mmxx
C and L intersect at the points A(x1, y1) andB(x2, y2).x1 and x2 are the roots of the quadraticequation
.0)191449(
)71(2)1(2
22
mm
xmmxm
(ii) From (b)(i),
2211
)71(2
m
mmxx
5 Equations of Circles
131
M(a, b) is the mid-point of AB.
2
2
21
1
)17(
1
)71(2
2
12
m
mm
m
mm
xxa
(c) M lies on L1 : x 1.
2
1or
3
1
0)12)(13(
016
17
11
)17(
1
2
22
2
mm
mm
mm
mmm
m
mm
a
35. (a) (i)
.....(2).018210:
......(1):22 yxyxC
mxyL
By substituting (1) into (2), we have
......(*)018)102()1(
018)(210)(22
22
xmxm
mxxmxx
L intersects C at two points A(x1, y1) andB(x2, y2).x1 and x2 satisfy (*).
221
2
221
1
181
1021
)102(
mxx
m
mm
mxx
22
2
22
22
222
2
212
212
21
)1(
)17107(4
)1(
)1(72100404
1
184
)1(
)102(
4)()(
m
mm
m
mmm
mm
m
xxxxxx
(ii)
2
2
221
2
221
221
221
221
2
1
)17107(4
))(1(
)()(
)()(
m
mm
xxm
mxmxxx
yyxxAB
(b) L is a tangent to C.
1or17
7
0)1)(717(
071017
017107
01
)17107(4
0
0
2
2
2
2
2
mm
mm
mm
mm
m
mm
AB
AB
36. (a) By substituting A(6, 3) into
0152: 22 yaxyxC , we have
4
0246
015)3(2)6(3)6( 22
a
a
a
By substituting A(6, 3) into L : y x b, we have
9
63
b
b
(b) (i)
.....(2)01524:
.....(1)9:22 yxyxC
xyL
By substituting (1) into (2), we have
(rejected)6or4
0)6)(4(
02410
048202
01518248118
015)9(24)9(
2
2
22
22
xx
xx
xx
xx
xxxxx
xxxx
By substituting x 4 into (1), we have
5
94
y
Coordinates of B )5,4(
(ii) Let S be the centre of C.
Coordinates of S
)1,2(
2
)2(,
2
4
Slope of SB 2)2(4
15
Slope of L12
1
The equation of L1 is
72
)]4([2
15
xy
xy
(c) Let D(h, k) be the intersection of L2 and C.The centre S of C is the mid-point of BD.
3and02
51and
2
42
kh
kh
Coordinates of D (0, 3)
Slope of L22
1ofslope 1 L
The equation of L2 is
32
xy
37. (a) Slope of L1 1)3(5
19
The equation of L1 is
4
)3(1
xy
xy
L2 L1
Slope of L2 1The equation of L2 is
8
)6(2
xy
xy
NSS Mathematics in Action 5A Full Solutions
132
(b) (i) Radius
50
)92()56( 22
CD
The equation of S is
010412
)50()2()6(22
222
yxyx
yx
(ii) Slope of the tangent to S parallel to L1
slope of L1
1Let y x c be the equation of the tangentto S parallel to L1.
)2......(010412
)1.......(22 yxyx
cxy
By substituting (1) into (2), we have
......(*)0)104()162(2
01044122
010)(412)(
22
222
22
ccxcx
cxxccxxx
cxxcxx
y x c is a tangent to S.For the equation (*),
6or14
0)6)(14(
0848
0848
020826416
02082)8(
0)104)(2(4)162(
0
2
2
22
22
22
cc
cc
cc
cc
cccc
ccc
ccc
The equations of the tangents to S parallel toL1 are y x 14 and y x 6.
38. (a)
3
1tan
3
1
)3(
1ofSlope 1
AOT
L
30AOT
(b) OT TA (tangent radius)i.e. 90ATO
4
230sin
sin
OAOA
OA
TAAOT
Coordinates of A = (4, 0)The equation of C1 is
4)4(
2)0()4(22
222
yx
yx
(c)
3
1
30tanofSlope 2
L
L2 passes through the origin O.The equation of L2 is
3
xy
(d) Radius of C1 = 2AB = 2Coordinates of B
0),2(
)0,24(
(e) Let D(h, 0) be the centre of C2 and r be the radius ofC2.
h
r30sin
h 2r
3
2
32
22
r
r
rr
rhOB
3
4
3
22
h
The equation of C2 is9
4
3
4 22
yx .
Alternative SolutionLet D(h, 0) be the centre of C2 and r be the radius ofC2.
ODU ~ OAT
rh
rhTA
UD
OA
OD
224
(corr. sides, ~ s)
3
2
32
22
r
r
rr
rhOB
3
4
3
22
h
The equation of C2 is9
4
3
4 22
yx .
39. (a) (i) Coordinates of R
1,2
2
2,
2
k
k
(ii) By substituting y = 0 into 03 cyx , wehave
3
003
cx
cx
Coordinates of P
0,
3
c
5 Equations of Circles
133
(b) (i) Slope of L 3)1(
3
L PR (tangent radius)
63
223
3
3
1
23
)1(03
1ofSlope
ck
kc
kc
PR
(ii) The circle C passes through
0,
3
cP .
By substituting
0,
3
cP and 6
3
2
ck
into C: x2 + y2 + kx + 2y – 5 = 0, we have
04518
0529
1
0529
2
9
05)0(23
63
20
3
2
2
22
22
cc
cc
ccc
ccc
(c) From (b)(ii),
15or3
0)15)(3(
045182
cc
cc
cc
When c = 3, 463
)3(2k
When c = 15, 463
)15(2k (rejected)
The equation of C is x2 + y2 – 4x + 2y – 5 = 0.
40. (a) (i) PQ and PS are tangents toC at Q and S respectively.PQ QR and PS SR tangent radius
i.e. PQR = PSR = 90
180
9090PSRPQR
PQRS is a cyclicquadrilateral. opp. s supp.
(ii) PQR = 90PR is a diameter of C1. (converse of in
semi-circle)
Coordinates of R
1),2(
2
)2(,
2
4
Centre of C1
2
15,
2
9
2
141,
2
112
ofpoint-mid PR
Radius of C1
2
169
4
169
4
169
2
1514
2
911
22
The equation of C1 is
08159
2
169
4
22515
4
819
2
169
2
15
2
9
22
22
22
yxyx
yyxx
yx
(b) (i)
)2(08159:
)1(0824:22
1
22
yxyxC
yxyxC
(2) – (1):
)3(
01313
xy
yx
By substituting (3) into (1), we have
1or4
0)1)(4(
043
0862
08)(24)(
2
2
22
xx
xx
xx
xx
xxxx
By substituting x = –4 into (3), we havey = –(–4) = 4
By substituting x = 1 into (3), we havey = –1Coordinates of Q 4),4(
Coordinates of S 1),1(
(ii) Slope of PQ3
2
)4(11
414
The equation of the tangent from P to C atQ is
3
20
3
2
)]4([3
24
xy
xy
Slope of PS2
3
111
)1(14
The equation of the tangent from P to C atS is
2
5
2
3
)1(2
3)1(
xy
xy
Multiple Choice Questions (p. 5.43)1. Answer: B
Centre
2
7,2
2
)7(,
2
4
NSS Mathematics in Action 5A Full Solutions
134
Radius
2
94
81
44
494
)4(2
7
2
422
2. Answer: CFor choice A,
radius
2
122
6
2
222
The equation represents an imaginary circle.For choice B,
x2 – y2 = 1The equation does not represent a circle.
For choice C,
02
12
01422
22
22
xyx
xyx
Radius
2
1
2
1
2
22
The equation represents a real circle.For choice D,
2x2 + y2 – 1 = 0The equation does not represent a circle.The answer is C.
3. Answer: B
Centre of C1
)1,3(
2
2,
2
6
Centre of C2 = (–3, –1)Let r be the radius of C2.The equation of C2 is
(*))1()3(
)]1([)]3([222
222
ryx
ryx
C2 passes through the origin.By substituting (0, 0) into (*), we have
10
)10()30(2
222
r
r
The equation of C2 is
026
10)1()3(22
22
yxyx
yx
4. Answer: A
Centre of C
4,2
2
)8(,
2
k
k
L divides the circle C into two equal parts.The centre of C lies on L.
7
07
05)4(32
2
k
k
k
5. Answer: A
Centre of C
4,2
5
2
8,
2
)5(
The centre of C lies in quadrant IV.
Radius of C
2
94
81
22
8
2
522
Distance between the centre of C and the origin
Cofradius4
89
)04(02
5 22
The origin lies outside the circle.The answer is A.
6. Answer: DC passes through (1, –2).By substituting (1, –2) intoC: x2 + y2 + ax – ay + b = 0, we have
(*)053
0)2()1()2(1 22
ba
baa
For I, a = 5 and b = 0 do not satisfy (*).For II, a = –2 and b = 1
3(–2) + 1 + 5 = 0They satisfy (*).
For III, a = –1 and b = –23(–1) + (–2) + 5 = 0
They satisfy (*).The answer is D.
7. Answer: DL: y = 2x – 1 (1)C1: x2 + y2 – 6x + a = 0 (2)C2: x2 + y2 + 4x + b = 0 (3)By substituting (1) into (2), we have
(*)0)1(105
06144
06)12(
2
22
22
axx
axxxx
axxx
5 Equations of Circles
135
L is a tangent to C1.For the equation (*),
4
02020100
0)1)(5(4)10(
02
a
a
a
By substituting (1) into (3), we have
(**)0)1(5
04144
04)12(
2
22
22
bx
bxxxx
bxxx
L intersects C2 at two points.For the equation (**),
1
01
0)1)(5(40
02
b
b
b
The answer is D.
8. Answer: D
)2(0652
)1(222
yxyx
cyx
From (1), we havey = 2x – c (3)
By substituting (3) into (2), we have
(*)0)65()124(5
06510244
06)2(52)2(
22
222
22
ccxcx
cxxccxxx
cxxcxx
L is a tangent to S.For the equation (*),
)0
(rejected2or3
0)2)(3(
06
02444
0120100201449616
0)65)(5(4)]124([
0
2
2
22
22
c
cc
cc
cc
cc
cccc
ccc
9. Answer: CBy substituting x = 3 into C: x2 + y2 – 3x – 4y – 5 = 0, wehave
1or5
0)1)(5(
054
054)3(332
22
yy
yy
yy
yy
A lies in quadrant IV.Coordinates of A = (3, –1)
Let S be the centre of C.
Coordinates of S
2,2
3
2
)4(,
2
)3(
Slope of SA 23
2
3)1(2
L1 SA (tangent radius)
Slope of L12
1
The equation of L1 is
052
322
)3(2
1)1(
yx
xy
xy
10. Answer: BLet x2 + y2 + Dx + Ey + F = 0 be the equation of the circle.
The circle passes through the origin, (a, 0) and (0, b).By substitution, we have
)3(
0)()0(0
)2(
0)0()(0
)1(0
0)0()0(00
2
22
2
22
22
bFbE
FbEDb
aFaD
FFaDa
F
FED
By substituting (1) into (2), we have
)0(
2
aaD
aaD
By substituting (1) into (3), we have
)0(
2
bbE
bbE
The equation of the circle is x2 + y2 – ax – by = 0.
11. Answer: ALet S be the centre of the circle.
Coordinates of S
2,
2
2
)(,
2
)(
ba
ba
Slope of OS
a
b
a
b
02
02
Slope of the tangentb
a
The tangent passes through the origin.The equation of the tangent is
0
byax
xb
ay