如何计算给定三点的圆的中心点？ [英] How to calculate the centre point of a circle given three points?

问题描述

我使用Javascript，我知道3点的位置。我希望用这些来找出一个圆圈的中心点。

我找到了这个逻辑（不是选择的答案，而是11个upvotes的答案）： https://math.stackexchange.com/questions/ 213658 /得到这个方程式的时候，给定3分

但我似乎无法得到我的头围绕着如何编写它的逻辑。

我不能使用边界框，这必须使用以下三点来完成：）

有什么想法？

解决方案

我最喜欢的分辨率：

将这三点转换为原点（减去（X0，Y0））。 / p>

通过两点和原点的圆的方程可以写成

  2X.Xc + 2Y.Yc =X²+Y²
  

在这两点上，你可以通过两个未知数得到一个两个方程组的简单系统，由Cramer

  Xc =（Z1.Y2 -  Z2.Y1）/ D 
 Yc =（X1.Z2  -  X2.Z1）/ D 
 
 D = 2（X1.Y2  -  X2.Y1），Z1 =X1²+Y1² ，Z2 =X2²+Y2²
  

将被翻译回来（添加（X0， Y0））。

---

当三点对齐时，公式失败由 D = 0 （或与分子相比较小）。

---

  X1- = X0; Y1- = Y0; X2- = X0; Y2- = Y0; 
 
 double Z1 = X1 * X1 + Y1 * Y1; 
 double Z2 = X2 * X2 + Y2 * Y2; 
 double D = 2 *（X1 * Y2-X2 * Y1）; 
 
 double Xc =（Z1 * Y2  -  Z2 * Y1）/ D + X0; 
 double Yc =（X1 * Z2-X2 * Z1）/ D + Y0; 
  

I am using Javascript and I know the positions of 3 points. I wish to use these to find out the center point of a circle.

I have found this logic (Not the chosen answer but the one with 11 upvotes) : https://math.stackexchange.com/questions/213658/get-the-equation-of-a-circle-when-given-3-points

But I can't seem to get my head around how to write the logic for it.

I can't use bounding box by the way, this has to be done using the three points :)

Any ideas ?

解决方案

My favorite resolution:

Translate the three points to bring one of them at the origin (subtract (X0,Y0)).

The equation of a circle through two points and the origin can be written

2X.Xc + 2Y.Yc = X² + Y²

Plugging the coordinates of the two points, you get an easy system of two equations in two unknowns, and by Cramer

Xc = (Z1.Y2 - Z2.Y1) / D
Yc = (X1.Z2 - X2.Z1) / D

D = 2(X1.Y2 - X2.Y1), Z1 = X1²+Y1², Z2 = X2²+Y2²

to be translated back (add (X0,Y0)).

---

The formula fails when the three points are aligned, which is diagnosed by D = 0 (or small in comparison to the numerators).

---

        X1-= X0; Y1-= Y0; X2-= X0; Y2-= Y0;

        double Z1= X1 * X1 + Y1 * Y1;
        double Z2= X2 * X2 + Y2 * Y2;
        double D= 2 * (X1 * Y2 - X2 * Y1);

        double Xc= (Z1 * Y2 - Z2 * Y1) / D + X0;
        double Yc= (X1 * Z2 - X2 * Z1) / D + Y0;

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