两边交叉区域 [英] Area of Intersection between Two Circles

问题描述

给定两个圈：

• C1 at（ x1 ，<$ c $


• C2（ x2 code>， y2 ）与 radius2



你如何计算交叉点的面积？当然，所有标准的数学函数（ sin ， cos 等）都是可用的。

解决方案

好的，使用Wolfram链接和Misnomer的提示来看等式14，我使用我列出的变量和这些中心（可以从中派生出来）：

  Double r = radius1; 
 Double R = radius2; 
 Double d = distance; 
 if（R  // swap 
 r = radius2; 
 R = radius1; （（d * d + r * r  -  R * R）/（2 * d * r））; b（b * b）
 Double part1 = r * r * Math.acos （（d * d + R * R-r * r）/（2 * d * R））;（b）b部分2 = R * R * Math.acos （d-r + R）*（d + r-R）*（d-r + R）*（d + r + R））; 
 
 Double intersectionArea = part1 + part2  -  part3; 
  

Given two circles:

• C1 at (x1, y1) with radius1
• C2 at (x2, y2) with radius2

How do you calculate the area of their intersection? All standard math functions (sin, cos, etc.) are available, of course.

解决方案

Okay, using the Wolfram link and Misnomer's cue to look at equation 14, I have derived the following Java solution using the variables I listed and the distance between the centers (which can trivially be derived from them):

Double r = radius1;
Double R = radius2;
Double d = distance;
if(R < r){
    // swap
    r = radius2;
    R = radius1;
}
Double part1 = r*r*Math.acos((d*d + r*r - R*R)/(2*d*r));
Double part2 = R*R*Math.acos((d*d + R*R - r*r)/(2*d*R));
Double part3 = 0.5*Math.sqrt((-d+r+R)*(d+r-R)*(d-r+R)*(d+r+R));

Double intersectionArea = part1 + part2 - part3;

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