已知圆c经过点A（0.-6）,B(2,-4)且圆心C在直线L:x-y+2=0上,求圆C的方程

let centre (x0,y0),radius r
equation of circle C:
(x-x0)^2+(y-y0)^2 =r^2
(x0,y0) is on L
x0-y0+2=0
x0=y0-2 (1)
A is on C
x0^2+(y0+6)^2 =r^2 (2)
B is on C
(x0-2)^2+(y0+4)^2 =r^2 (3)
(1)=(2)
x0^2+(y0+6)^2 = (x0-2)^2+(y0+4)^2 (4)
sub (1) into (4)
(y0-2)^2+(y0+6)^2 = (y0-4)^2+(y0+4)^2
8y0+40=32
y0=-1
x0=-3
r^2=34
equtaion of circle
(x+3)^2+(y+1)^2=34