若x^2+y^2-2x=0,则y^2-x^2-4x的取值范围为?

x^2+y^2-2x=0
y^2=2x-x^2
R=y^2-x^2-4x=2x-x^2-x^2-4x=-2x^2-2x=-2(x+0.5)^2+1
又因为y^2=2x-x^2 >=0 所以0<=x<=2
故-12<=R<=0

yuan gen shuang qu xian de guan xi

x^2+y^2-2x=0
(x-1)^2+y^2=1

令x=1+cosa，y=sina

y^2-x^2-4x
=sina^2-(1+cosa)^2-4(1+cosa)
=1-cosa^2-1-2cosa-cosa^2-4-4cosa
=-2cosa^2-6cosa-4
=-2(cosa+3/2)^2+1/2

1/2<=cosa+3/2<=5/2

1/4<=(cosa+3/2)^2<=25/4

y^2-x^2-4x的取值范围为:[-12,0].

tohomelin 的是简单的正解~~