# Distance between centers of two intersecting circles if the radii and common chord length is given

• Last Updated : 05 Nov, 2021

Given are two circles, with given radii, which intersect each other and have a common chord. The length of the common chord is given. The task is to find the distance between the center of the two circles.

Examples:

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

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

• let the length of common chord AB = x
• Let the radius of the circle with center O is OA = r2
• Radius of circle with center P is AP = r1
• From the figure, OP is perpendicular AB
AC = CB
AC = x/2 (Since AB = x)
• In triangle ACP,
AP^2 = PC^2+ AC^2 [By Pythagoras theorem]
r1^2 = PC^2 + (x/2)^2
PC^2 = r1^2 – x^2/4
• Consider triangle ACO
r2^2 = OC^2+ AC^2[By Pythagoras theorem]
r2^2 = OC^2+ (x/2)^2
OC^2 = r2^2 – x^2/4
• From the figure, OP = OC + PC
OP = √( r1^2 – x^2/4 ) + √(r2^2 – x^2/4)

Distance between the centers = 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)

Below is the implementation of the above approach:

## 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`
Output:
```distance between the centers is 44
```

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