已知圆的方程：x^2+y^2-2(2m-1)x+2(m+1)y+5m^2-2m-2=0

、已知圆的方程x^2+y^2-2(2m-1)x+2(m+1)y+5m^2-2m-2=0,求圆心的轨迹?[x-(2m-1)]^2+[y+(m+1)]^2=(2m-1)^2+(m+1)^2-(5m^2-2m-2),[x-(2m-1)]^2+(y+(m+1)]^2=4,圆心x=2m-1,y=-m-1,x+2y=-3,圆心的轨迹是直线x+2y=-3.所以...