设u=x-y,则x=u+y,代入x^2+y^2-4x=0
(u+y)^2+y^2-4(u+y)=0
u^2+2uy+y^2+y^2-4u-4y=0
2y^2+(2u-4)y+u^2-4u=0
判别式=(2u-4)^2-8(u^2-4u)>=0
4u^2-16u+16-8u^2+32u>=0
4u^2-16u-16