已知圆C:x²+y²-2x+4y-4=0,问是否在斜率为1直线l,使以l被圆C截得的弦AB为直径的圆原点,求出直线L的方程

已知圆C:x²+y²-2x+4y-4=0,问是否在斜率为1直线l,使以l被圆C截得的弦AB为直径的圆原点,求出直线L的方程

x²+y²-2x+4y-4=0
(x - 1)² + (y + 2)² = 9
C(1,-2),半径R = 3
设AB的中点为D,新圆半径为r
CD斜率为 -1,方程为:y + 2 = -(x - 1),y = -x -1
直线l:y = x + b,x - y + b = 0
交点D(-(b + 1)/2,(b - 1)/2)
AD = r = OD,r² = (b + 1)²/4 + (b - 1)²/4
AC² = AD² + CD² = r² + CD²
9 = (b + 1)²/4 + (b - 1)²/4 + [1 + (b + 1)/2]² + [-2 - (b -1)/2]²
(b + 4)(b - 1) = 0
b = -4,y = x - 4
或b = 1,y = x + 1