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

Given Centre 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) coordinates, 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 coordinates     cout << "x2 = "             << (float)(2 * c1 - x1)<< "  ";           // find end point for y coordinates     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 coordinates         System.out.print( "x2 = "                             + (2 * c1 - x1) + " ");                   // find end point for y coordinates         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.

## Python3

 # 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 coordinates     print("x2 =", (2 * c1 - x1), end=" ")           # find end point for y coordinates     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 coordinates         Console.Write("x2 = "+ (2 * c1 - x1) + " ");                   // find end point for y coordinates         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

 

## Javascript

 

Output

x2 = 10 y2 = 11

Time complexity: O(1)

Auxiliary Space: O(1)
Similarly if we given a centre (c1, c2) and other end coordinate (x2, y2) of a diameter and we finding a (x1, y1) coordinates

 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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