已知圆C：x^2+y^2-2x+4y+4=0,是否存在斜率1的直线l,使以l被圆C所截得的弦AB为直径的圆经过原点?

let line l be:
y = x + c (1)
C:x^2+ y^2-2x +4y + 4 =0 (2)
let l cut C at A(x1,y1),B(x2,y2)
|AB|^2 = (x1-x2)^2 + (y1-y2)^2 (3)
Sub (1) into (2)
x^2 + (x+c)^2 - 2x+ 4(x+c) + 4 =0
2x^2 + 2(c+1)x + c^2+4c+4 =0
x1+x2 = -(c+1)
x1x2 = (c^2+4c+4)/2
(x1-x2)^2 = (x1+x2)^2 - 2x1x2
= (c+1)^2 - (c^2+4c+4)
= -2c-3 (4)
Similarly,we have
x^2+ y^2-2x +4y + 4 =0
(y-c)^2 +y^2 -2(y-c) + 4y +4 =0
2y^2 -2(c-1)y+c^2+2c+4=0
y1+y2 = c-1
y1y2= (c^2+2c+4)/2
(y1-y2)^2 = (y1+y2)^2 - 2y1y2
= (c-1)^2 - (c^2+2c+4)
= -4c-3 (5)
centre of the circle C'
= mid point of AB
= ((x1+x2)/2 ,(y1+y2)/2)
=( -(c+1)/2,(c-1)/2 )
if the cicle pass through O(0,0),then
|C'O| = √ {(c^2+1)/2}
Sub (4),(5) into (3)
|AB|^2 = -2c-3 +(-4c-3)
= -6c -6
= -6(c+1)
|C'O| = (1/2) |AB|
2(c^2+1) = -6(c+1)
2c^2+6c+8 =0
c^2+3c+4 =0
△= 9 - 16 < 0
no such c exists
不存在斜率1的直线l,使以l被圆C所截得的弦AB为直径的圆经过原点

设直线的方程为：y=x+b,且圆的方程经整理得：(x-1)^2+(y+2)^2=1,两式联立得：2x^2+4x+2b+8=0,所以：x1+x2=-2,y1+y2=2b-2,所以另一圆的圆心为（-1,b-1），所以两圆心的斜率是：