题目内容:
过点A(1,-1),B(-1,1)且圆心在直线x+y-2=0上的圆的方程是( )
A.
(x-3)2+(y+1)2=4 A.(x-3)2+(y+1)2=4
B. (x+3)2+(y-1)2=4
B. (x+3)2+(y-1)2=4
C. (x-1)2+(y-1)2=4
C. (x-1)2+(y-1)2=4
D. (x+1)2+(y+1)2=4
D. (x+1)2+(y+1)2=4
参考答案:
答案解析: