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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