急! F5 Equations of Circles 8q4

23a) Radius of C1 = (1/2)√(42 + 22 + 4 x 59) = 8

Centre of C1 is (-2, 1)

b) Distance between the centres = √[(1 + 2)2 + (5 - 1)2] = 5

Hence radius of C2 should be 8 - 5 = 3 since they touch internally.

Equation of C2:

(x - 1)2 + (y - 5)2 = 9

x2 + y2 - 2x - 10y + 17 = 0

24a) Centre of C1 is (4, -5)

Radius of C1 is (1/2)√(82 + 102 - 4 x 16) = 5

Hence radius of C2 is 2 since C2 is smaller than C1

Equation of C2 is:

(x - 4)2 + (y + 5)2 = 4

x2 + y2 - 8x + 10y + 25 = 0

b) With area ratio = 4:1, radius ratio = 2:1. Hence radius of C2 is 5/2

Equation of C2 is:

(x - 4)2 + (y + 5)2 = 25/4

4x2 + 4y2 - 32x + 40y + 139 = 0

26a) Sub y = 0 into C:

x2 + 10x + 16 = 0

(x + 8)(x + 2) = 0

Hence A is (-8, 0) and B is (-2, 0)

Sub x = 0 into C:

y2 - 8y + 16 = 0

(y - 4)2 = 0

Hence C is (0, 4)

bi) Slope of BC = 2

Hence equation of AD is y/(x + 8) = 2

2x - y + 16 = 0

ii) Sub y = 2x + 16 into C:

x2 + (2x + 16)2 + 10x - 8(2x + 16) + 16 = 0

5x2 + 58x + 144 = 0

(5x + 18)(x + 8) = 0

x = -8 or -18/5

Hence D is at (-18/5, 44/5)