从三个点中找出圆心的算法是什么？ [英] What is the algorithm for finding the center of a circle from three points?

问题描述

我在圆周上有三个点：

pt A =（A.x，A.y）;
pt B =（B.x，B.y）;
pt C =（Cx，Cy）;

我如何计算圆心？

我找到了答案并实现了一个工作解决方案：

  pt circleCenter（pt A，pt B，pt C）{
 
 float yDelta_a = By  -  Ay; 
 float xDelta_a = B.x  -  A.x; 
 float yDelta_b = C.y  -  B.y; 
 float xDelta_b = C.x  -  B.x; 
 pt center = P（0,0）; 
 
 float aSlope = yDelta_a / xDelta_a; 
 float bSlope = yDelta_b / xDelta_b; 
 center.x =（aSlope * bSlope *（Ay-Cy）+ bSlope *（Ax + Bx）
  -  aSlope *（B.x + Cx））/（2 *（bSlope-aSlope） ）; 
 center.y = -1 *（center.x  - （A.x + B.x）/ 2）/ aSlope +（A.y + B.y）/ 2; 
 
 return center; 
} 
  

解决方案

深度计算。这里有一个简单的步骤： http://paulbourke.net/geometry/circlesphere/。一旦你有圆的方程，你可以简单地把它放在一个涉及H和K的形式中。（h，k）点将成为中心。

（在链接上向下滚动以获得方程式）

I have three points on the circumference of a circle:

pt A = (A.x, A.y); pt B = (B.x, B.y); pt C = (C.x, C.y);

How do I calculate the center of the circle?

Implementing it in Processing (Java).

I found the answer and implemented a working solution:

 pt circleCenter(pt A, pt B, pt C) {

    float yDelta_a = B.y - A.y;
    float xDelta_a = B.x - A.x;
    float yDelta_b = C.y - B.y;
    float xDelta_b = C.x - B.x;
    pt center = P(0,0);

    float aSlope = yDelta_a/xDelta_a;
    float bSlope = yDelta_b/xDelta_b;  
    center.x = (aSlope*bSlope*(A.y - C.y) + bSlope*(A.x + B.x)
        - aSlope*(B.x+C.x) )/(2* (bSlope-aSlope) );
    center.y = -1*(center.x - (A.x+B.x)/2)/aSlope +  (A.y+B.y)/2;

    return center;
  }

解决方案

It can be a rather in depth calculation. There is a simple step-by-step here: http://paulbourke.net/geometry/circlesphere/. Once you have the equation of the circle, you can simply put it in a form involving H and K. The point (h,k) will be the center.

(scroll down a little ways at the link to get to the equations)

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