点到圆的最近距离公式推导

最新推荐文章于 2022-04-11 13:27:35 发布

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最新推荐文章于 2022-04-11 13:27:35 发布

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文章标签：点到圆的距离公式

本文链接：https://blog.youkuaiyun.com/pg3412/article/details/41964061

该距离公式在 Circle Fitting相关的一篇文章中用到，现实现其推导过程：

设圆的一般的方程形式： $a_{1}(x^{2}+y^{2})+a_{2}x+a_{3}y+a_{4}=0$ ，任一点P的坐标 $(x_{i},y_{i})$ ，点P到圆上点得最短距离 $d_{i}$ 的公式？

推导过程：

（1）由圆一般方程形式可以推导出圆的标准方程形式：

$(x_{i}+\frac{a_{2}}{2a_{1}})^{2}+(y_{i}+\frac{a_{3}}{2a_{1}})^{2}=\frac{a_{2}^{2}+a_{3}^{2}-4a_{1}a_{4}}{(2a_{1})^{2}}$ ，且有： $r^{2}=\frac{a_{2}^{2}+a_{3}^{2}-4a_{1}a_{4}}{(2a_{1})^{2}}$ ，即： $a_{2}^{2}+a_{3}^{2}-4a_{1}a_{4}\geqslant 0$

圆心坐标： $(-\frac{a_{2}}{2a_{1}},-\frac{a_{3}}{2a_{1}})$ ；半径： $r$

（2）点到圆的最短距离 $d_{i}$ ：

点到圆的最短距离等于点到圆心的距离减去半径的绝对值；

那么点在圆外（点在圆内时取绝对值即可）：

$d_{i}=\sqrt{(x_{i}+\frac{a_{2}}{2a_{1}})^{2}+(y_{i}+\frac{a_{3}}{2a_{1}})^{2}}-r\\ =\frac{\sqrt{(2a_{1}x_{i}+a_{2})^{2}+(2a_{1}y_{i}+a_{3})^{2}}-\sqrt{a_{2}^{2}+a_{3}^{2}-4a_{1}a_{4}}}{2a_{1}}\\ =\frac{[\sqrt{(2a_{1}x_{i}+a_{2})^{2}+(2a_{1}y_{i}+a_{3})^{2}}-\sqrt{a_{2}^{2}+a_{3}^{2}-4a_{1}a_{4}}][\sqrt{(2a_{1}x_{i}+a_{2})^{2}+(2a_{1}y_{i}+a_{3})^{2}}+\sqrt{a_{2}^{2}+a_{3}^{2}-4a_{1}a_{4}}]}{2a_{1}[\sqrt{(2a_{1}x_{i}+a_{2})^{2}+(2a_{1}y_{i}+a_{3})^{2}}+\sqrt{a_{2}^{2}+a_{3}^{2}-4a_{1}a_{4}}]}\\ =\frac{(2a_{1}x_{i}+a_{2})^{2}+(2a_{1}y_{i}+a_{3})^{2}-(a_{2}^{2}+a_{3}^{2}-4a_{1}a_{4})}{2a_{1}[\sqrt{(2a_{1}x_{i}+a_{2})^{2}+(2a_{1}y_{i}+a_{3})^{2}}+\sqrt{a_{2}^{2}+a_{3}^{2}-4a_{1}a_{4}}]}=(1)//$

将 $(2a_{1}x_{i}+a_{2})^{2}+(2a_{1}y_{i}+a_{3})^{2}$ 展开：

$(2a_{1}x_{i}+a_{2})^{2}+(2a_{1}y_{i}+a_{3})^{2}=4a_{1}^{2}(x_{i}^{2}+y_{i}^{2})+4a_{1}a_{2}x_{i}+4a_{1}a_{3}y_{i}+a_{2}^{2}+a_{3}^{2}$ 代入（1）可得：

$//(1)=\frac{[4a_{1}^{2}(x_{i}^{2}+y_{i}^{2})+4a_{1}a_{2}x_{i}+4a_{1}a_{3}y_{i}+a_{2}^{2}+a_{3}^{2}]- (a_{2}^{2}+a_{3}^{2}-4a_{1}a_{4})}{2a_{1}[\sqrt{4a_{1}(x_{i}^{2}+y_{i}^{2})+4a_{1}a_{2}x_{i}+4a_{1}a_{3}y_{i}+a_{2}^{2}+a_{3}^{2}}+\sqrt{a_{2}^{2}+a_{3}^{2}-4a_{1}a_{4}}]}\\ =\frac{2(a_{1}(x_{i}^{2}+y_{i}^{2})+a_{2}x_{i}+a_{3}y_{i}+a_{4})}{\sqrt{4a_{1}(x_{i}^{2}+y_{i}^{2})+4a_{1}a_{2}x_{i}+4a_{1}a_{3}y_{i}+a_{2}^{2}+a_{3}^{2}}+\sqrt{a_{2}^{2}+a_{3}^{2}-4a_{1}a_{4}}}=(2)//$