a-maths!!!!!

a)通過C1和C2交點的圓族:
x²+y²-2x-3+k(x²+y²-4x-4y+7) = 0 {k是不等於-1的實數}
x2+y2-2x-3+kx2+ky2-4kx-4ky+7k = 0
(k+1)x2+(k+1)y2+(-2-4k)x-4ky+(7k-3) = 0
x2+y2+(-2-4k)x/(k+1) – 4ky/(k+1) + (7k-3)/(k+1) = 0
圓心: [ ( 1 + 2k ) / ( k + 1 ) , 2k / ( k + 1 ) ], 所以
3( 1 + 2k ) / ( k + 1 )+8k / ( k + 1 )-1= 0
3 + 6k + 8k – k – 1 = 0
13k = -2
k = -2/13
於是,
x²+y²-2x-3+(-2/13)(x²+y²-4x-4y+7) = 0
11x2+11y2-18x+8y-53 = 0
b)圓的半徑: ( 1 / 2 )( √(-2-4k)2/(k+1)2 + (-4k)2/(k+1)2 – 4 (7k–3) / ( k+1) )
= [( √(k2 + 4 )) / ( k + 1 )]
圓心離x+2y-1=0的距離 = 半徑
d = l (Ax1 + By1 + C ) / √A2 + B2 l
d = l [( 1 ) ( 1 + 2k ) / ( k + 1 ) + 4k / ( k + 1 ) – 1 ] / √12+22 l
= l √5k / ( k + 1 ) l
於是,
l √5k / ( k + 1 ) l = ( √(k2 + 8k + 4 )) / ( k + 1 )
√5k = √( k2 + 4 )
5k2 = k2 + 4
4k2 = 4
k = 1 或 – 1 (捨去)
所以,
x²+y²-2x-3+x²+y²-4x-4y+7 = 0
2x2 + 2y2 – 6x – 4y + 4 = 0
x2 + y2 – 3x – 2y + 2 = 0