已知直线l:x+2y-3=0与圆C：x^2+y^2+x-2cy+c=0的两个交点为A、B,且以AB为直径的圆过坐标原点,

设A(x1,y1),B(x2,y2),则AB直径的圆过原点表明∠ACB = 90°,于是AC⊥BC,用斜率表示即
y2/x2 * y1/x1 = -1,
即
x1*x2 + y1*y2 = 0
将x = 3 - 2y代入方程得
(3-2y)^2+ y^2 +(3-2y) - 2cy + c = 0.
整理得
5y^2 -(14+2c)y + 12 + c = 0.
所以y1 + y2 = 14/5+ 2/5 * c,y1*y2 = 12/5+1/5 * c
x1*x2 + y1*y2 = (3-2y1)(3-2y2) + y1 * y2
= 9 - 6(y1 + y2) + 5y1*y2
= 9 - 6(14/5 + 2/5*c) + 5(12/5 + 1/5*c)
=21/5 - 7/5*c
得c = 3
发现自己算错好多,改了好多次……

把圆方程C：x^2+y^2+x-2cy+c=0写成标准式：
(x-1/2)^2+(y-c)^2=(c-1/2)^2
直线l:x+2y-3=0代入上式得2个分别关于x,y的2次方程：
(3-2y-1/2)^2+(y-c)^2=(c-1/2)^2
(x-1/2)^2+(-x/2+3/2-c)^2=(c-1/2)^2
化简得：
2y^2-2(c+5)y+c...