x^2+y^2+Dx+Ey+F=0(D^2+E^2-4F＞0)
(x+D/2)^2+(y+E/2)^2=D^2/4+E^2/4-F
曲线关于x+y=0对称,圆心在x=-y上
故x=-y即x=-D/2,y=-E/2
-D/2=E/2
D/2+E/2=0
D+E=0