题目

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) 答案：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.

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