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

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 = 40Output:44Input:r1 = 14, r2 = 7, x = 10Output: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 <bits/stdc++.h>`

`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