方程AX^2+AY^2+DX+EY+F=0中,若A不等于0,怎么才不为圆求详解

Question

方程AX^2+AY^2+DX+EY+F=0中,若A不等于0,怎么才不为圆
求详解

Answer 1

AX^2+AY^2+DX+EY+F=0
A(X+D/2A)^2+A(Y+E/2A)^2+F-(D^2+E^2)/4A=0
(X+D/2A)^2+(Y+E/2A)^2=(D^2+E^2-4AF)/4A^2
不为圆
(D^2+E^2-4AF)/4A^2≤0
D^2+E^2-4AF≤0