求圆C:x^2+y^2-4x+2y+1=0关于直线x-y+1=0对称的方程

(x-2)^2+(y+1)^2-5+1=0
(x-2)^2+(y+1)^2=4
(2,-1) R=2
设（2,-1）关于直线x-y+1=0对称的点为（x,y)
中点（（x+2)/2,(y-1)/2)
k=(y+1)/(x-2)
(x+2)/2-（y-1)/2+1=0
(y+1)/(x-2）=-1
y+1=2-x
y=1-x
(x+2)-(y-1)+2=0
x+2-(-x)+2=0
x+2+x+2=0
2x=-4
x=-2
y=1-(-2)=3
(-2,3)
(x+2)^2+(y-3)^2=4