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数学
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圆x^2+y^2-2y*cosθ-sin^2θ=0的半径是
人气:363 ℃ 时间:2020-06-26 09:16:05
解答
1
x^2+y^2-2y*cosθ-sin^2θ=0
==>x^2+y^2-2y*cosθ+cos^2θ-cos^2θ-sin^2θ=0
==>x^2+y^2-2y*cosθ+cos^2θ=cos^2θ+sin^2θ=1
==>x^2+(y-cosθ)^2=1
即圆心为(0,cosθ),半径为1
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