Given Center coordinate (c1, c2) and one coordinate (x1, y1) of the diameter of a circle, find the other end coordinate point (x2, y2) of diameter.

Examples:

Input : x1 = –1, y1 = 2, and c1 = 3, c2 = –6 Output : x2 = 7, y2 = -14 Input : x1 = 6.4, y1 = 3 and c1 = –10.7, c2 = 4 Output : x2 = -27.8, y2 = 5

**The Midpoint Formula:**

The midpoint of two ends coordinates points, (x1, y2) and (x2, y2) is the point M can be found by using:

We have need of a (x2, y2) cordinates so we apply the midpoint the formula

c_{1}= ((x_{1}+x_{2})/2), c2 = ((y_{1}+y_{2})/2) 2*c_{1}= (x_{1}+x_{2}), 2*c2 = (y_{1}+y_{2})x_{2}= (2*c_{1}- x_{1}), y_{2}= (2*c_{2}- y_{1})

## C++

`// CPP program to find the ` `// other-end point of diameter` `#include <iostream>` `using` `namespace` `std;` ` ` `// function to find the` `// other-end point of diameter` `void` `endPointOfDiameterofCircle(` `int` `x1,` ` ` `int` `y1, ` `int` `c1, ` `int` `c2)` `{` ` ` `// find end point for x cordinates` ` ` `cout << ` `"x2 = "` ` ` `<< (` `float` `)(2 * c1 - x1)<< ` `" "` `;` ` ` ` ` `// find end point for y cordinates` ` ` `cout << ` `"y2 = "` `<< (` `float` `)(2 * c2 - y1);` ` ` `}` `// Driven Program` `int` `main()` `{` ` ` `int` `x1 = -4, y1 = -1;` ` ` `int` `c1 = 3, c2 = 5;` ` ` ` ` `endPointOfDiameterofCircle(x1, y1, c1, c2);` ` ` ` ` `return` `0;` `}` |

## Java

`// Java program to find the other-end point of ` `// diameter` `import` `java.io.*;` ` ` `class` `GFG {` ` ` ` ` `// function to find the other-end point of ` ` ` `// diameter` ` ` `static` `void` `endPointOfDiameterofCircle(` `int` `x1,` ` ` `int` `y1, ` `int` `c1, ` `int` `c2)` ` ` `{` ` ` ` ` `// find end point for x cordinates` ` ` `System.out.print( ` `"x2 = "` ` ` `+ (` `2` `* c1 - x1) + ` `" "` `);` ` ` ` ` `// find end point for y cordinates` ` ` `System.out.print(` `"y2 = "` `+ (` `2` `* c2 - y1));` ` ` `}` ` ` ` ` `// Driven Program` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `x1 = -` `4` `, y1 = -` `1` `;` ` ` `int` `c1 = ` `3` `, c2 = ` `5` `;` ` ` ` ` `endPointOfDiameterofCircle(x1, y1, c1, c2);` ` ` `}` `}` ` ` `// This code is contributed by anuj_67.` |

## Python 3

`# Python3 program to find the ` `# other-end point of diameter` ` ` `# function to find the` `# other-end point of diameter` `def` `endPointOfDiameterofCircle(x1, y1, c1, c2):` ` ` ` ` `# find end point for x cordinates` ` ` `print` `(` `"x2 ="` `, (` `2` `*` `c1 ` `-` `x1), end` `=` `" "` `)` ` ` ` ` `# find end point for y cordinates` ` ` `print` `(` `"y2 ="` `, (` `2` `*` `c2 ` `-` `y1))` ` ` `# Driven Program` `x1 ` `=` `-` `4` `y1 ` `=` `-` `1` `c1 ` `=` `3` `c2 ` `=` `5` ` ` `endPointOfDiameterofCircle(x1, y1, c1, c2)` ` ` `# This code is contributed by Smitha.` |

## C#

`// C# program to find the other - ` `// end point of diameter` `using` `System;` `class` `GFG {` ` ` ` ` `// function to find the other - end` ` ` `// point of diameter` ` ` `static` `void` `endPointOfDiameterofCircle(` `int` `x1,` ` ` `int` `y1,` ` ` `int` `c1, ` ` ` `int` `c2)` ` ` `{` ` ` `// find end point for x cordinates` ` ` `Console.Write(` `"x2 = "` `+ (2 * c1 - x1) + ` `" "` `);` ` ` ` ` `// find end point for y cordinates` ` ` `Console.Write(` `"y2 = "` `+ (2 * c2 - y1));` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `Main ()` ` ` `{` ` ` `int` `x1 = -4, y1 = -1;` ` ` `int` `c1 = 3, c2 = 5;` ` ` ` ` `endPointOfDiameterofCircle(x1, y1, c1, c2);` ` ` `}` `}` ` ` `// This code is contributed by anuj_67.` |

## PHP

`<?php` `// PHP program to find the ` `// other-end point of diameter` ` ` `// function to find the` `// other-end point of diameter` `function` `endPointOfDiameterofCircle(` `$x1` `,` ` ` `$y1` `, ` `$c1` `, ` `$c2` `)` `{` ` ` `// find end point for x cordinates` ` ` `echo` `"x2 = "` `,(2 * ` `$c1` `- ` `$x1` `),` `" "` `;` ` ` ` ` `// find end point for y cordinates` ` ` `echo` `"y2 = "` `, (2 * ` `$c2` `- ` `$y1` `);` `} ` ` ` `// Driven Program` `$x1` `= -4;` `$y1` `= -1;` `$c1` `= 3;` `$c2` `= 5;` ` ` `endPointOfDiameterofCircle(` `$x1` `, ` `$y1` `,` ` ` `$c1` `, ` `$c2` `);` ` ` `// This code is contributed by Smitha` `?> ` |

**Output**

x2 = 10 y2 = 11

**Similarly if we given a center (c1, c2) and other end codrdinate (x2, y2) of a diameter and we finding a (x1, y1) cordinates**

Proof for (x1, y1) :c1 = ((x1+x2)/2), c2 = ((y1+y2)/2) 2*c1 = (x1+x2), 2*c2 = (y1+y2) x1 = (2*c1 - x2), y1 = (2*c2 - y2)

**So The other end coordinates (x1, y1) of a diameter is**

x1 = (2*c1 - x2), y1 = (2*c2 - y2)

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