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Find the center of the circle using endpoints of diameter

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  • Last Updated : 16 Jun, 2022

Given two endpoint of diameter of a circle (x1, y1) and (x2, y2) find out the center of a circle. 
Examples : 
 

Input  : x1 = -9, y1 = 3, and 
         x2 = 5, y2 = –7
Output : -2, –2

Input  :  x1 = 5, y1 = 3 and 
          x2 = –10 y2 = 4
Output : –2.5, 3.5

 

Midpoint Formula: 
The midpoint of two points, (x1, y2) and (x2, y2) is : M = ((x 1 + x 2) / 2, (y 1 + y 2) / 2) 
The center of the circle is the mid point of its diameter so we calculate the mid point of its diameter by using midpoint formula. 
 

center of the circle using endpoints of diameter

 

C++




// C++ program to find the
// center of the circle
#include <iostream>
using namespace std;
 
// function to find the
// center of the circle
void center(int x1, int x2,
            int y1, int y2)
{
     
    cout << (float)(x1 + x2) / 2 <<
          ", " << (float)(y1 + y2) / 2;
}
 
// Driven Program
int main()
{
    int x1 = -9, y1 = 3, x2 = 5, y2 = -7;
    center(x1, x2, y1, y2);
    return 0;
}

Java




// Java program to find the
// center of the circle
class GFG {
     
    // function to find the
    // center of the circle
    static void center(int x1, int x2,
                            int y1, int y2)
    {
         
        System.out.print((float)(x1 + x2) / 2
            + ", " + (float)(y1 + y2) / 2);
    }
     
    // Driver Program to test above function
    public static void main(String arg[]) {
         
        int x1 = -9, y1 = 3, x2 = 5, y2 = -7;
        center(x1, x2, y1, y2);
    }
}
 
// This code is contributed by Anant Agarwal.

Python3




# Python3 program to find
# the center of the circle
 
# Function to find the
# center of the circle
def center(x1, x2, y1, y2) :
 
    print(int((x1 + x2) / 2), end= "")
    print(",", int((y1 + y2) / 2) )
 
# Driver Code
x1 = -9; y1 = 3; x2 = 5; y2 = -7
center(x1, x2, y1, y2)
 
# This code is contributed by Smitha Dinesh Semwal

C#




// C# program to find the
// center of the circle
using System;
 
class GFG {
     
    // function to find the
    // center of the circle
    static void center(int x1, int x2,
                            int y1, int y2)
    {
         
        Console.WriteLine((float)(x1 + x2) / 2
                + ", " + (float)(y1 + y2) / 2);
    }
     
    // Driver Program to test above function
    public static void Main() {
         
        int x1 = -9, y1 = 3, x2 = 5, y2 = -7;
        center(x1, x2, y1, y2);
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to find the
// center of the circle
 
// function to find the
// center of the circle
function center($x1, $x2, $y1, $y2)
{
     
    echo((float)($x1 + $x2) / 2 . ", " .
                (float)($y1 + $y2) / 2);
}
 
// Driven Code
$x1 = -9; $y1 = 3; $x2 = 5; $y2 = -7;
center($x1, $x2, $y1, $y2);
 
// This code is contributed by Ajit.
?>

Javascript




<script>
 
// javascript program to find the
// center of the circle
 
// function to find the
    // center of the circle
    function center(x1, x2,
                     y1, y2)
    {
           
        document.write((x1 + x2) / 2
            + ", " + (y1 + y2) / 2);
    }
 
// Driver Function
 
         let x1 = -9, y1 = 3, x2 = 5, y2 = -7;
        center(x1, x2, y1, y2);
     
    // This code is contributed by susmitakundugoaldanga.
</script>

Output : 
 

-2, -2

Time complexity: O(1)
Auxiliary Space: O(1) 


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