在最接近给定点的圆上找到点的最佳方法 [英] Best way to find a point on a circle closest to a given point

问题描述

给定一个点 (pX, pY) 和一个圆心 (cX,cY) 和半径 (r)(pX, pY) ?

Given a point (pX, pY) and a circle with a known center (cX,cY) and radius (r), what is the shortest amount of code you can come up with to find the point on the circle closest to (pX, pY) ?

我有一些代码可以工作，但它涉及将圆转换为形式为 (x - cX)^2 + (y - cY)^2 = r^2 的方程(其中 r 是半径)和使用点 (pX, pY) 到 (cX, cY) 的直线方程来创建要求解的二次方程.

I've got some code kind of working but it involves converting the circle to an equation of the form (x - cX)^2 + (y - cY)^2 = r^2 (where r is radius) and using the equation of the line from point (pX, pY) to (cX, cY) to create a quadratic equation to be solved.

一旦我解决了它会做的错误，但这似乎是一个不优雅的解决方案.

Once I iron out the bugs it'll do, but it seems such an inelegant solution.

推荐答案

其中 P 是点，C 是中心，R 是半径，用合适的数学"语言:

where P is the point, C is the center, and R is the radius, in a suitable "mathy" language:

V = (P - C); Answer = C + V / |V| * R;

在哪里 |V|是V的长度.

where |V| is length of V.

好的，好的

double vX = pX - cX;
double vY = pY - cY;
double magV = sqrt(vX*vX + vY*vY);
double aX = cX + vX / magV * R;
double aY = cY + vY / magV * R;

易于扩展到 >2 维.

easy to extend to >2 dimensions.

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