圆心C在直线l:x+2y=0,圆C过点A(2,-3),且截直线m:x-y-1=0所得弦长为2√2,求圆C的方程

因为圆心在直线x+2y=0上,所以设圆C坐标是（-2m,m),半径为R
R^2=(m+3)^2+(-2m-2)^2=m^2+6m+9+4m^2+8m+4=5m^2+14m+13
圆心C到直线X-Y-1=0的距离d=|-2m-m-1|/根号2=|3m+1|/根号2
根据勾股定理 得：
R^2=d^2+[(2根号2)/2]^2
5m^2+14m+13=[|3m+1|/根号2]^2+2=(9m^2+6m+1)/2+2
10m^2+28m+26=9m^2+6m+1+4
m^2+22m+21=0
(m+1)(m+21)=0
m=-1或m=-21
R^2=5-14+13=4或R^2=5*21^2-14*21+13=1924
圆心坐标是：（2,-1）或(42,-21)
那么圆方程是：（x-2)^2+(y+1)^2=4或(x-42)^2+(y+21)^2=1924