{注：形如x^2+y^2+Dx+Ey+F＝0
即(x-D/2)^2+(y-E/2)^2＝[D^2+E^2-4F]/4
當(dāng)半徑R＝[D^2+E^2-4F]/4＞0
則x^2+y^2+Dx+Ey+F＝0的圖像為圓；反之,則不是圓}
∵k1、k2≠-1 ∴k1+1、k2+1≠0
故x^2+y^2+[4(k-1)x]/(k+1)+[(4k-2)y]/(k+1)+(4-8k)/(k+1)＝0
當(dāng){[4(k-1)/(k+1)]^2+[(4k-2)/(k+1)]^2-4(4-8k)/(k+1)}/4＞0
則方程對(duì)應(yīng)的曲線即為圓
求解的工作就留給樓主了.
注：僅供參考!