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© Oxford Fajar Sdn. Bhd. (008974-T) 2012

CHAPTER 5 ANALYTIC GEOMETRY

Focus on Exam 5

1 x2 + y2 – 4x – 6y + 4 = 0Centre: C

1 is (2, 3)

Radius: r1 = 22 + 32 – 4 = 3

x2 + y2 – 10x – 14y + 70 = 0Centre: C

1 is (5, 7)

Radius: r2 = 52 + 72 – 70 = 2

C2C1

3 2P

Distance C1C

2 = 32 + 42 = 5

Since C1C

2 = r

1 + r

2, the circles touch

externally. Coordinates of point of contact,

P = 1(3)(5) + (2)(2)3 + 2

, (3)(7) + (2)(3)

3 + 2 2 = 119

5,

275 2 = (3.8, 5.4)

2 x2 + y2 + 4y + 3 = 0Centre: C

1 is (0, –2)

Radius: r1 = 4 – 3 = 1

x2 + y2 – 12x – 12y – 49 = 0Centre: C

2 is (6, 6)

Radius: r2 = 62 + 62 + 49 = 11

C2C1

1 9P(x, y)

Distance C1C

2 = 62 + 82 = 10

Since C1C

2 = r

2 ± r

1, the circles touch

internally.

0 = (1)(6) + 10x

11 ⇒ x = –

610

= –0.6

–2 = (1)(6) + 10y

11 ⇒ y = –

2810

= –2.8

p = (–0.6, –2.8)

3 x2 + y2 – 2x – 2y – 14 = 0Centre: C

1 is (1, 1)

Radius: r1 = 12 + 12 + 14 = 4

x2 + y2 – 3x – 24 = 0

Centre: C2 is 13

2, 02

Radius: r2 = 13

222

+ 24 = 1052

5.12

C1

C2

Distance C1C

2 = 11

222

+ 12 = 52

1.11

Since C1C

2 < r

2 – r

1,

circles C1 lies entirely inside C

2.

4

O

C (r, r)

y

x

r

r (8, 4)

Chap-5.indd 1 3/1/2012 11:01:20 AM


ACE AHEAD Mathematics (T) First Term2

Equation of the circle is of the form (x – r)2 + (y – r)2 = r 2

i.e. x2 + y 2 – 2rx – 2ry + r 2 = 0Since the circle passes through (8, 4)64 + 16 – 16r – 8r + r 2 = 0

r 2 – 24r + 80 = 0(r – 4)(r – 20) = 0

r = 4 or 20Equation of the circles are

x2 + y 2 – 8x – 8y + 16 = 0and x2 + y 2 – 40x – 40y + 400 = 0

5

O C

y

x

Let the equation of the circle be x2 + y2 + 2gx + c = 0

( f = 0 since the centre lies on the x-axis)

Since (1, 1) lies on the circle,1 + 1 + 2g + c = 0 …(1)

Since (4, –2) lies on the circle,16 + 4 + 8g + c = 0 …(2)

(2) – (1),18 + 6g = 0 ⇒ g = –3

From (1),2 – 6 + c = 0 ⇒ c = 4

Equation of the circle is x2 + y 2 – 6x + 4 = 0

6

O

C(r, r)

y

x

2r

Centre: (r, r)

Radius: 2r

Equation of circle is of the form (x – r)2 + (y – r)2 = 2r2

i.e. x2 + y2 – 2rx – 2ry = 0Since (3, –1) lies on the circle,9 + 1 – 6r + 2r = 0

10 = 4r

r = 52

Equation of circle is x2 + y 2 – 5x – 5y = 0

7

O

C(r, 3)

y

x

r(0, 3)

A B

Let the radius of the circle be r. ∴ Coordinates of centre C is (r, 3)Equation of circle is

(x – r)2 + (y – 3)2 = r 2

i.e. x2 + y2 – 2rx – 6y + 9 = 0The point (8, 7) lies on the circle.∴ 64 + 49 – 16r – 42 + 9 = 0

16r = 80r = 5

Equation of the circle isx2 + y2 – 10x – 6y + 9 = 0

At the x-axis, y = 0x2 – 10x + 9 = 0(x – 1)(x – 9) = 0 ⇒ x = 1 or 9

Gradient of CA = 34

⇒ grad. of tangent = – 43

Equation of tangent at A is

y = – 43

(x – 1) …(1)

Gradient of CB = – 34

⇒ grad. of tangent = 43

Equation of tangent at B is

y = 43

(x – 9) …(2)

Solving (1) and (2) for point of intersection,

– 43

(x – 1) = 43

(x – 9)

1 – x = x – 910 = 2x ⇒ x = 5

From (1), y = – 43

(5 – 1) = – 163

Point of intersection is 15, – 163 2

8 Let the equation of the circle be x2 + y 2 + 2gx + 2 f y + c = 0

Point (10, 6) 100 + 36 + 20g + 12f + c = 0 …(1)

Chap-5.indd 2 3/1/2012 11:01:23 AM


Fully Worked Solution 3

Point (–4, 4) 16 + 16 – 8g + 8f + c = 0 …(2)Point (4, 8) 16 + 64 + 8g + 16f + c = 0 …(3)(1) – (2), 104 + 28g + 4f = 0 …(4)(3) – (2), 48 + 16g + 8f = 0 …(5) –(5) + 2 × (4),

160 + 40g = 0 ⇒ g = – 4From (5), 48 – 64 + 8f = 0 ⇒ f = 2From (2),

32 + 32 + 16 + c = 0 ⇒ c = –80Equation of circle is

x2 + y 2 – 8x + 4y – 80 = 0Substitute x = 10, y = –10,100 + 100 – 80 – 40 – 80 = 0[ P(10, –10) lies on the circle.

PQ(4 , –2)C(x, y)

Centre of given circle is Q(4, –2)

Radius of given circle is 42 + 22 + 80 = 10Let the centre of the circle be C(x, y)

1 3

P C Q

By the ratio theorem,

x = (1)(4) + 3(10)

4 = 34

4 = 17

2

y = (1)(–2) + 3(–10)

4 = –8

[ equation of circle is

1x – 172 2

2

+ (y + 8)2 = 254

x2 – 17x + 2894

+ y 2 + 16y + 64 = 254

x2 + y 2 – 17x + 16y + 130 = 0

9

P

C

(x – 5)2 + (y – 7)2 = 25

Centre of circle C is (5, 7)

Gradient of PC = 7 – 35 – 2

= 43

Gradient of tangent at P = – 34

Equation of tangent at P is

y – 3 = – 34

(x – 2)

4y – 12 = –3x + 63x + 4y = 18

10

C

A(6, –1)B(1, 0)

2x − 3y

= 153x + 2y = 3

Let the centre of the circle be C.

Gradient of line 2x – 3y = 15 is 23

.

[ gradient of BC = 23

Equation of BC is

y = 23

(x – 1) …(1)

Gradient of line 3x + 2y = 3 is – 32

.

[ gradient of AC = – 32

Equation of AC is

y + 1 = – 32

(x – 6) … (2)

Solving (1) and (2),

23

(x – 1) = – 32

(x – 6) – 1

4x – 4 = –9x + 54 – 613x = 52

x = 4, y = 2[ C(4, 2)

Radius of circle BC = 32 + 22

= 13Equation of circle is

(x – 4)2 + (y – 2)2 = 13x2 + y 2 – 8x – 4y + 7 = 0

Chap-5.indd 3 3/1/2012 11:01:25 AM



11

C

P(0, 8)3y

− 4x

− 24

= 0

3x − y = 7

Gradient of line 3y – 4x – 24 = 0 is 43

[ gradient of PC = – 34

Equation of PC is

y – 8 = – 34

(x – 0)

i.e. 4y + 3x = 32 …(1)line: 3x – y = 7 …(2)

Solving: 5y = 25 ⇒ y = 5, x = 4, ∴ C(4, 5)

Radius of circle = 42 + 32 = 5

Equation of circle is (x – 4)2 + (y – 5)2 = 52

x2 + y 2 – 8x – 10y + 16 = 0

12 Let the equation of the circle be x2 + y 2 + 2gx + 2f y + c = 0

(2, 5): 4 + 25 + 4g + 10f + c = 0 …(1)(4, 3): 16 + 9 + 8g + 6f + c = 0 …(2)x + y = 3: –g – f = 3 …(3)(1) – (2): 4 – 4g + 4f = 0i.e. –g + f = –1 …(4)(4) – (3): 2f = – 4

f = –2g = –1

From (1): 29 – 4 – 20 + c = 0c = –5

Equation of circle is x2 + y2 – 2x – 4y – 5 = 0

Centre is (1, 2), radius = 10

13 (x – 3)2 + (y – 4)2 = 25x2 – 6x + 9 + y2 – 8y + 16 = 25

x2 + y2 – 6x – 8y = 0Since the coefficient of x2 is the same as the coefficient of y 2 and there is no term in xy, the above equation represents a circle.Since x = 0, y = 0 satisfies the equation, the circle passes through the origin.

O

y

x

C

P

RQ

Centre of circle is C(3, 4)Coordinates of R are (6, 8)

At P, y = 0x2 – 6x = 0

x(x – 6) = 0x = 0 or 6 ⇒ P(6, 0)

At Q, x = 0y2 – 8y = 0

y(y – 8) = 0y = 0 or 8 ⇒ Q(0, 8)

Gradient of PQ = 8–6

= – 43

Gradient of CP = – 43

Gradient of PQ = gradient of CPC lies on PQ ⇒ PQ is a diameter

Gradient of OC = 43

Gradient of tangent at O = – 36

Equation of tangent is

y = – 34

x or 3x + 4y = 0

14 Equation of the circle is(x – 1)(x – 7) + (y – 0)(y – 8) = 0x2 + y 2 – 8x – 8y + 7 = 0

At the x-a xis, y = 0x2 – 8x + 7 = 0

(x – 1)(x – 7) = 0x = 1 or 7, A(1, 0), B(7, 0)

At the y-axis, x = 0y2 – 8y + 7 = 0

(y – 1)(y – 7) = 0y = 1 or 7, C(0, 1), D(0, 7)

AB = 6 units CD = 6 units ⇒ AB = CDCentre of circle C is (4, 4)

Gradient of CA = 43

Chap-5.indd 4 3/1/2012 11:01:27 AM



Gradient of tangent at A = – 34

Equation of tangent at A is

y = – 34

(x – 1) …(1)

Gradient of BC = 4–3

Gradient of tangent at B = 34

Equation of tangent at B is

y = 34

(x – 7) …(2)

Solving (1) and (2),

– 34

(x – 1) = 34

(x – 7)

1 – x = x – 7 8 = 2x ⇒ x = 4

x = 4, y = – 94

T14, – 942

) 1 7 4 1 )0 0 – 94

0

Area of D ABT = 12

)71– 942 – 1– 9

42)= 1

2 )– 63

4 + 9

4)= 1

2 154

4 2= 27

4 unit2

= 634

unit2

15

Ox

y

(–3, 2)

(–7, 6)

From a sketch, the parabola opens to the left. Equation is of the form

( y – k)2 = – 4a(x – h)∴ ( y – 2)2 = – 4a(x + 3)(–7, 6): 42 = – 4a(– 4)

a = 1Equation: (y – 2)2 = – 4(x + 3)

16 y2 + 2y – 6x + 25 = 0 y2 + 2y = 6x – 25 y2 + 2y + 1 = 6x – 24

( y + 1)2 = 6(x – 4)Comparing with (y – k)2 = 4a(x – h)

4a = 6 ⇒ a = 32

Vertex (4, –1); Focus 1112

, –12

Ox

y

(4, –1)

17 4x2 + 9y2 – 16x = 04(x2 – 4x) + 9y2 = 0

4[(x – 2)2 – 4] + 9y2 = 04(x – 2)2 + 9y2 = 16(x – 2)2

4 + 9y2

16 = 1

[ a = 2, b = 43

Centre of ellipse is (2, 0)Length of major axis = 2a = 4

Length of minor axis = 2b = 83

Ox

y

18 25x2 + 9y2 –150x – 18y + 9 = 025(x2 – 6x) + 9(y2 – 2y) + 9 = 0

25[(x – 3)2 – 9] + 9[(y – 1)2 – 1] + 9 = 0 25(x – 3)2 + 9(y – 1)2 = 225

(x – 3)2

9 + (y – 1)2

25 = 1

The equation represents an ellipse with vertical major axis

a = 5, b = 3Centre (3, 1); Vertices (3, –4), (3, 6)

Chap-5.indd 5 3/1/2012 11:01:33 AM



Ox

y

19 4y2 – 16x2 – 16y – 48 = 0 y2 – 4x2 – 4y – 12 = 0

(y2 – 4y) – 4x2 = 12 (y – 2)2 – 4x2 = 16

(y – 2)2

16 – x2

4 = 1

Hyperbola with vertical transverse axis

Centre: (0, 2); Vertices: (0, –2), (0, 6)

Asymptotes: y – 2 = ± 2(x – 0)y = ± 2x + 2

x

y

O

20 4x2 – 9y2 + 8x + 54y – 113 = 04(x2 + 2x) – (9y2 – 6y) = 113

4(x2 + 2x + 1) – 9( y2 – 6y + 9) = 113 + 4 – 814(x + 1)2 – 9(y – 3)2 = 36

(x + 1)2

9 – ( y – 3)2

4 = 1

Hyperbola with horizontal transverse axis

Centre: (–1, 3); Vertices: (2, 3), (– 4, 3)

Slopes = ± 23

Ox

y

21 y2 = 16x

y = 8x – 4 ⇒ x = y + 4

8

Solving:

y2 = 16( y + 4)8

y2 – 2y – 8 = 0(y – 4)(y + 2) = 0

y = 4 or –2

y = 4, x = 1 and y = –2, x = 14

Points of intersection are

(1, 4) and 114

, –22

22 Let the equation of the tangent be y = 2x + c …(1)x2 + 2y 2 = 8 …(2)

Solving: x2 + 2(2x + c)2 = 8

x2 + 8x2 + 8cx + 2c2 – 8 = 09x2 + 8cx + 2c2 – 8 = 0

Roots are equal ⇒64c2 – 4(9)(2c2 – 8) = 0

16c2 – 9(2c2 – 8) = 016c2 – 18c2 + 72 = 0

2c2 = 72c2 = 36c = ±6

Equation of tangents are y = 2x ± 6

23 x = t(t – 2) … (1)

y = 2(t – 1) ⇒ t = y2

+ 1

Substitute into (1):

x = 1y2

+ 12 1y2

+ 1 – 22x = 1y

2 + 12 1y

2 – 12

x = y 2

4 – 1

y2 = 4(x + 1) – Cartesian equation

Ox

y

2

–2

–1

Chap-5.indd 6 3/1/2012 11:01:35 AM



24 x = 2t1 + t2

⇒ x2 = 4t2

(1 + t2)2

y = 1 – t2

1 + t2 ⇒ y2 = (1 – t2)2

(1 + t2)2

x2 + y2 = 4t2 + (1 – t2)2

(1 + t2)2

= t4 + 2t2 + 1(1 + t2)2

= (1 + t2)2

(1 + t2)2

= 1The Cartesian equation x2 + y2 = 1 represents a circle with centre at origin and radius of 1 unit.

Ox

y

1

1

25 x = t + 1 …(1)

y = 12

t2 – 1 …(2)

From (1): t = x – 1Substitute into (2):

y = 12

(x – 1)2 – 1

2( y + 1) = (x – 1)2

This equation represents a parabola with

vertex (1, –1), focus 11, – 122.

Ox

y

26 x = 2(t2 – 4)t

…(1)

y = t4 + 16

t2 …(2)

Squaring (1), x2 = 4(t4 – 8t2 + 16)t2

= 41t4 + 16

t2 – 82

x2 = 4(y – 8)Vertex (0, 8), Focus (0, 9)

Ox

y

8

27 x = 4 cos q + 1

⇒ cos q = x – 14

y = 3 sin q – 2

⇒ sin q = y + 2

3Using sin2 q + cos2 q = 1(y + 2)2

9 +

(x – 1)2

16 = 1

This is an equation of an ellipse with horizotal major axis,a = 4, b = 3 c2 = a2 – b2 = 42 – 32 = 7c = 7

Centre: (1, –2); Foci: (1 ± 7 , –2)

Ox

y

28 x = 3 cos q – 2 ⇒ cos q = x + 23

y = 4 sin q + 1 ⇒ sin q = y – 14

Using sin2 q + cos2 q = 1,

( y – 1)2

16 + (x + 2)2

9 = 1

a = 4, b = 3, c2 = a2 – b2 = 7 ⇒ c = 7Centre: (–2, 1); Foci: (–2, 1 ± 7 )

Chap-5.indd 7 3/1/2012 11:01:38 AM



Ox

y

29 x = 21t + 1t 2 ⇒ t + 1

t = x

2

t2 + 2 + 1t2

= x2

4 …(1)

y = 121t – 1

t 2 ⇒ t – 1t = 2y

t2 – 2 + 1t2

= 4y 2 …(2)

(1) – (2), 4 = x2

4 – 4y 2

x2

16 – y 2 = 1

This equation represents a hyperbola with a horizotal transverse axis a = 4, b = 1

c2 = a2 + b2 = 16 + 1 = 17 ⇒ c = 17

Vertices: (±4, 0) Foci: (± 17 , 0)

Asymptotes: y = ± 16

x

Ox

y

30 x = tan q – 1 ⇒ tan q = x + 1

y = 2 sec q + 1 ⇒ sec q = y – 1

2Using sec2 q – tan2 q = 1

(y – 1)2

4 – (x + 1)2 = 1

Hyperbola with a vertical transverse axisa = 2, b = 1c2 = a2 + b2 = 4 + 1 = 5 ⇒ c = 5

Centre: (–1, 1) Foci: (–1, 1 ± 5 )

Ox

y

Chap-5.indd 8 3/1/2012 11:01:40 AM

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