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楼主 |
发表于 2020-11-2 23:52
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\(\because\;{\small a\cos\theta+b\sin\theta+c=}(\frac{a}{2r}{\small+r\cos\theta})^2{\small+}(\frac{b}{2r}{\small+r\sin\theta})^2{\small+}\frac{4r^2(c-r^2)-(a^2+b^2)}{4r^2}\)
\(\qquad\)当且仅当\(\,c>\sqrt{a^2+b^2}\,\)时存在\(\;\;r=\small\sqrt{\frac{1}{2}(c\pm\sqrt{c^2-(a^2+b^2)})}>0\,\)使
\(\qquad\sqrt{a\cos\theta+b\sin\theta+c}=\big|r(\cos\theta,\sin\theta)-(\frac{-a}{2r},\frac{-b}{2r})\big|\,\)为平面两点距离. |
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