A-maths!!!!!!!!

a) i)common chord : C1 – C2 = 0
x²+y²+2x-4y+1 – (x²+y²-4x-10y+19) =0
6x + 6y -18 = 0
i.e. x + y -3 = 0

ii)
Family of circles passes through A and B:
x²+y²+2x-4y+1 + k (x + y -3) = 0
x²+y² +(2+k)x + (k-4)y +(1 -3k) = 0

Let the centre of circle C be G.
G lies on AB, so G[-(2+k)/2, -(k-4)/2] satisfied line AB.
x + y -3 = 0
-(2+k)/2 -(k-4)/2 – 3 = 0
2+k + k-4 +6 = 0
2k = -4
k = -2
∴eq : x²+y² +(2+k)x + (k-4)y +(1 -3k) = 0
x²+y² -6y +7 = 0

b)
Centre G: (0,3)
rC3 = | (x + y -3)/√(1²+1²)|
= |(0+3-3)/√2|
= 0
i.e.
eq: x² - (y-3)² = 0

2007-06-07 15:21:43 補充：
b)部份有更正：Center G is the same as the centre of C1, but not the smallest circle. Sorry for that.So, G [-2/2, -(-4)/2] = (-1, 2)rC3 = | (x y -3)/√(1² 1²)|= |(-1 2-3)/√2|= 2/√2= √2i.e. eq: (x 1)² (y-2)² = (√2)²

C1:x^2+y^2+2x-4y+1=0 ------------- (1)
C2:x^2+y^2-4x-10y+19=0 -------------- (2)

ai)
(1) - (2), 6x + 6y - 18 = 0
x + y - 3 = 0

ii)
x = 3 - y ------- (3)
sub (3) into (1), and 代簡, y^2 - 6y + 8 = 0
(y - 2)(y - 4) = 0
y = 2, x = 1
y = 4, x = -1
A = (-1, 4), B = (1, 2)
centre = (0, 3)
radius = sqrt(1^2 + 1^2) = sqrt(2)
x^2 + (y - 3)^2 = 2


b)
centre = (-1,2)
radius = ! -1 + 2 -3 ! / sqrt(2) = sqrt(2)
(x + 1)^2 + (y - 2)^2 = 2