Question

圆C₁:x²+y²+x+2y+=0与圆C₂:x²+y²-2xsinθ-2y+sin²θ=0相外切,则θ的值是(    )

A.-                                        B.

C.2kπ-(k∈Z)                         D.2kπ-或2kπ+(k∈Z)

Answer 1

D

解析：圆C₁:(x+)²+(y+1)²=1.

所以C₁(-,-1),r₁=1.

圆C₂:(x-sinθ)²+(y-1)²=1.

所以C₂(sinθ,1),r₂=1.

因为⊙C₁与⊙C₂外切,

所以|C₁C₂|=r₁+r₂,

即=2,得sinθ=-.

所以θ=2kπ-或2kπ+,k∈Z.故选D.