Question

已知圆C:x²+y²-2x+4y=0，则过原点且与圆C相切的直线方程为

A.y=-2x                                       B.y=x

C.y=x                                      D.y=2x

Answer 1

答案:C

解析：圆C:(x-1)²+(y+2)²=5,圆心C(1,-2)，半径5.

∵直线过原点，∴设直线方程为y=kx,即kx-y=0.

∵直线与圆相切，∴=,得k=,则直线为y=x.