已知圆C1:(x+1)2+(y−1)2=1,圆C2与圆C1关于直线x-y=0对称,则圆C2的方程为( ) A.(x-1)2+(y+1)2=1 B.(x-1)2+(y-1)2=1 C.(x+1)2+(y+1)2=1 D.(x+1)2+(y-
问题描述:
已知圆C1:(x+1)2+(y−1)2=1,圆C2与圆C1关于直线x-y=0对称,则圆C2的方程为( )
A. (x-1)2+(y+1)2=1
B. (x-1)2+(y-1)2=1
C. (x+1)2+(y+1)2=1
D. (x+1)2+(y-1)2=1
答
∵圆C1:(x+1)2+(y−1)2=1,
∴圆C1的圆心C1(-1,1),半径r1 =1,
∵圆C2与圆C1关于直线x-y=0对称,
∴圆C2的圆心C2(1,-1),半径r2=1,
∴圆C2的方程为(x-1)2+(y+1)2=1.
故选:A.