Very Urgent .. 2000CE..amaths..thanks

(a)(i)
substitute k=-1 into F
x^2 + y^2 + (4k+4) x + (3k+1) y - (8k+8) = 0
x^2 + y^2 + (4(-1)+4) x + (3(-1)+1) y - (8(-1)+8) = 0
x^2 + y^2 -2y = 0
This is the circle C1
So C1 is a circle in F
(ii)
sub y=0, then x^2=0 x=0
There are only one point (0,0) on the x-axis
So C1 touches the x - axis
(b)(i)
sub y=0 in F
x^2 + y^2 + (4k+4) x + (3k+1) y - (8k+8) = 0
We have
x^2 + (4k+4) x - (8k+8) = 0
Since the discriminant =0
[4(k+1)]^2+4*8(k+1)=0
16(k+1)^2+32(k+1)=0
(k+1)^2+2(k+1)=0
(k+1)(k+3)=0
k=-1 or -3
So the equation of C2 is
x^2 + y^2 + (4(-3)+4) x + (3(-3)+1) y - (8(-3)+8) = 0
x^2 + y^2 - 8x - 8y + 16 = 0
(ii)
The radius of C1 is 1/2√(2^2)=1

The radius of C2 is 1/2√(8^2+8^2-4*16)=4
The center of C1 (0,1)
The center of C2 (4,4)
Distance between center of C1 and C2 is
√(4^2+3^2)=5

The radius of C1 +The radius of C2=Distance between center of C1 and C2
So C1 and C2 touch externally