# Find the other-end coordinates of diameter in a circle

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


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

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


## C++

 // CPP program to find the   // other-end point of diameter  #include  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

 

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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