39
NSS Mathematics in Action 5A Full Solutions 98 5 Equations of Circles Review Exercise 5 (p. 5.5) 1. 0 8 ) 7 )( 1 ( 4 6 2 The equation x 2 + 6x + 7 = 0 has two distinct real roots. 2. 0 ) 8 )( 18 ( 4 ) 24 ( 2 The equation 18x 2 24x + 8 = 0 has one double real root. 3. The graph of 3 6 2 x kx y has no x-intercepts. i.e. The equation 0 3 6 2 x kx has no real roots. 3 36 12 0 ) 3 )( ( 4 6 0 2 k k k 4. 0 8 2 0 3 3 y x y x ) 2 ( ) 1 ( (1) 2 + (2) : 2 0 14 7 x x By substituting x = 2 into (1), we have 3 0 3 ) 2 ( 3 y y The solution of the simultaneous equations is (2, 3). 5. x y y x 4 0 4 3 4 2 ) 2 ( ) 1 ( By substituting (2) into (1), we have 1 or 4 0 ) 1 )( 4 ( 0 4 3 2 y y y y y y By substituting y = 4 into (1), we have 4 0 4 ) 4 ( 3 4 x x By substituting y = 1 into (1), we have 4 1 0 4 ) 1 ( 3 4 x x The solutions of the simultaneous equations are (4, 4) and 1 , 4 1 . 6. The equation of L is i.e. 0 6 2 4 2 2 ) 4 ( 2 1 1 y x x y x y 7. Slope of L 3 2 4 2 ) 1 ( 3 The equation of L is i.e. 0 5 3 2 8 2 3 3 ) 4 ( 3 2 ) 1 ( y x x y x y 8. (a) Slope of L 1 k k 3 ) ( 3 Slope of L 2 3 3 4 3 4 3 k k L 1 L 2 2 4 2 3 4 1 3 3 4 3 1 of slope of Slope 2 1 k k k k k k L L (b) (2) 0 3 3 2 : (1) 0 24 2 3 : 2 1 y x L y x L (1) 3 + (2) 2: 6 0 78 13 x x By substituting x = 6 into (1), we have 3 0 24 2 ) 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 1 2 2 2 2 2 y x y x (b) The equation of the circle is i.e. 5 ) 3 ( ) 2 ( ) 5 ( )] 3 ( [ ) 2 ( 2 2 2 2 2 y x y x Quick Practice 5.2 (p. 5.7) (a) 2 2 2 2 2 5 ) 1 ( )] 5 ( [ 25 ) 1 ( ) 5 ( y x y x Centre ) 1 , 5 ( , radius 5 (b) 2 2 2 2 2 2 2 ) 8 ( )] 1 ( [ ) 1 ( 8 )] 1 ( [ ) 1 ( 16 ) 1 ( 2 ) 1 ( 2 y x y x y x Centre ) 1 , 1 ( , radius ) 2 2 (or 8

5 Equations of Circles · 25 0 2 7 2 5 2 2 ð> ... Let P(x, y) be a point on the circle apart from A and B. ... The equation of the circle is x ð+y2 ð+x ð-8y ð=0

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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

qq

q

q

qq

qq

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

NSS Mathematics in Action 5A Full Solutions

136

12. Answer: BLet S be the centre of the circle C and T be the intersectionof L1 and C.

Coordinates of S

0),5(

0,2

)10(

OS = 5

Radius of C

5

202

102

TS

52

20

5)5( 222

222

OT

OT

OSTSOT (Pyth. theorem)

2

152

5

tan

OT

TS