急! F5 Equations of Circles 8q5

27a) Sub y = 1:

x2 - 2x - 15 = 0

(x - 5)(x + 3) = 0

x = -3 or 5

Hence A is (-3, 1) and B is (5, 1)

b) Mid-point of AB is (-1 ,1) which is the centre, and radius = 4

Equation of C2 is:

(x + 1)2 + (y - 1)2 = 16

x2 + y2 + 2x - 2y - 14 = 0

c) Centre of C1 is (1, -2)

Sub x = 1 and y = -2 into equation of C2:

12 + (-2)2 + 2 - 2(-2) - 14 =/= 0

Hence centre of C1 does not lie on C2.

28a) Sub y = 4 - x into C:

x2 + (4 -x)2 + 4x - 2(4 -x) - 20 = 0

2x2 + 2x - 12 = 0

x2 + x - 6 = 0

(x + 3)(x - 2) = 0

x = -3 or 2

Hence A is (-3, 7) and B is (2, 2)

bi) Centre of C is (-2, 1).

Slope of line joining centre of C and A is -6, hence tangent at A has a slope of 1/6, with equation:

(y - 7)/(x + 3) = 1/6

6y - 42 = x + 3

x - 6y + 45 = 0

Slope of line joining centre of C and B is 1/4, hence tangent at B has a slope of -4, with equation:

(y - 2)/(x - 2) = -4

y - 2 = 8 - 4x

4x + y - 10 = 0

30) By the fact that radius perp. to chord bisects chord, the x-coordinate of the centre is -5.

Also since it touches the y-axis, radius = 5

Let the centre be (-5, k), then

(-5 + 1)2 + k2 = 25 (Distance between centre and Q is 5)

k2 = 9

k = 3 (since the centre should have positive y-coordinate)

So equation is:

(x + 5)2 + (y - 3)2 = 25

x2 + y2 + 10x - 6y + 9 = 0