Skip to content
Related Articles

Related Articles

Improve Article
Find the other-end coordinates of diameter in a circle
  • Last Updated : 09 Jun, 2021

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:
M = \frac{x_{1} + x_{2}}{2}, \ \ \frac{y_{1} + y_{2}}{2}

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 <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 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




<?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 coordinates
    echo "x2 = ",(2 * $c1 - $x1)," ";
     
    // find end point for y coordinates
    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
?>

Javascript




<script>
 
// Javascript 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 coordinates
    document.write("x2 = " + (2 * c1 - x1) + "  ");
     
    // Find end point for y coordinates
    document.write("y2 = " + (2 * c2 - y1));
     
}
 
// Driver code
let x1 = -4, y1 = -1;
let c1 = 3, c2 = 5;
 
endPointOfDiameterofCircle(x1, y1, c1, c2);
 
// This code is contributed by jana_sayantan   
     
</script>

Output 

x2 = 10 y2 = 11

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)

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live




My Personal Notes arrow_drop_up
Recommended Articles
Page :