如果给出半径和共和弦长度，则两个相交圆的中心之间的距离

给定的是两个具有给定半径的圆，它们彼此相交并具有相同的弦。给出了普通和弦的长度。任务是找到两个圆心之间的距离。

例子：

Input:  r1 = 24, r2 = 37, x = 40
Output: 44

Input: r1 = 14, r2 = 7, x = 10
Output: 17

方法：

• 令普通和弦的长度AB = x
• 设以O为中心的圆的半径为OA = r2
• 中心为P的圆的半径为AP = r1
• 从图中可以看出， OP垂直于AB
  AC = CB
  AC = x / 2 (因为AB = x)
• 在三角形ACP中，
  AP ^ 2 = PC ^ 2 + AC ^ 2 [根据毕达哥拉斯定理]
  r1 ^ 2 = PC ^ 2 +(x / 2)^ 2
  PC ^ 2 = r1 ^ 2 – x ^ 2/4
• 考虑三角ACO
  r2 ^ 2 = OC ^ 2 + AC ^ 2 [根据毕达哥拉斯定理]
  r2 ^ 2 = OC ^ 2 +(x / 2)^ 2
  OC ^ 2 = r2 ^ 2 – x ^ 2/4
• 从图中， OP = OC + PC
  OP =√(r1 ^ 2 – x ^ 2/4)+√(r2 ^ 2 – x ^ 2/4)

  Distance between the centres = sqrt((radius of one circle)^2 – (half of the length of the common chord )^2) + sqrt((radius of the second circle)^2 – (half of the length of the common chord )^2)

  下面是上述方法的实现：

  C++

  // C++ program to find
  // the distance between centers
  // of two intersecting circles
  // if the radii and common chord length is given
    
  #include 
  using namespace std;
    
  void distcenter(int r1, int r2, int x)
  {
      int z = sqrt((r1 * r1)
                   - (x / 2 * x / 2))
              + sqrt((r2 * r2)
                     - (x / 2 * x / 2));
    
      cout << "distance between the"
           << " centers is "
           << z << endl;
  }
    
  // Driver code
  int main()
  {
      int r1 = 24, r2 = 37, x = 40;
      distcenter(r1, r2, x);
      return 0;
  }

  Java

  // Java program to find
  // the distance between centers
  // of two intersecting circles
  // if the radii and common chord length is given
  import java.lang.Math; 
  import java.io.*;
    
  class GFG {
        
  static double distcenter(int r1, int r2, int x)
  {
      double z = (Math.sqrt((r1 * r1)
                  - (x / 2 * x / 2)))
              + (Math.sqrt((r2 * r2)
                  - (x / 2 * x / 2)));
    
      System.out.println ("distance between the" +
                          " centers is "+ (int)z );
      return 0;
  }
    
  // Driver code
  public static void main (String[] args) 
  {
      int r1 = 24, r2 = 37, x = 40;
      distcenter(r1, r2, x);
  }
  }
    
  // This code is contributed by jit_t.

  Python3

  # Python program to find
  # the distance between centers
  # of two intersecting circles
  # if the radii and common chord length is given
    
  def distcenter(r1, r2, x):
      z = (((r1 * r1) - (x / 2 * x / 2))**(1/2)) +\
      (((r2 * r2)- (x / 2 * x / 2))**(1/2));
    
      print("distance between thecenters is ",end="");
      print(int(z));
    
  # Driver code
  r1 = 24; r2 = 37; x = 40;
  distcenter(r1, r2, x);
    
  # This code has been contributed by 29AjayKumar

  C#

  // C# program to find
  // the distance between centers
  // of two intersecting circles
  // if the radii and common chord length is given
  using System;
    
  class GFG
  {
            
  static double distcenter(int r1, int r2, int x)
  {
      double z = (Math.Sqrt((r1 * r1)
                  - (x / 2 * x / 2)))
              + (Math.Sqrt((r2 * r2)
                  - (x / 2 * x / 2)));
    
      Console.WriteLine("distance between the" +
                          " centers is "+ (int)z );
      return 0;
  }
    
  // Driver code
  static public void Main ()
  {
      int r1 = 24, r2 = 37, x = 40;
      distcenter(r1, r2, x);
  }
  }
    
  // This code is contributed by jit_t

  输出：

  distance between the centers is 44