Distance between Incenter and Circumcenter of a triangle using Inradius and Circumradius

Namachila November 17, 2025

0 1 2 minutes read

Given two integers r and R representing the length of Inradius and Circumradius respectively, the task is to calculate the distance d between Incenter and Circumcenter.

Inradius The inradius( r ) of a regular triangle( ABC ) is the radius of the incircle (having center as l), which is the largest circle that will fit inside the triangle.
Circumradius: The circumradius( R ) of a triangle is the radius of the circumscribed circle (having center as O) of that triangle.

Examples:

Input: r = 2, R = 5
Output: 2.24

Input: r = 5, R = 12
Output: 4.9

Approach:
The problem can be solved using Euler’s Theorem in geometry, which states that the distance between the incenter and circumcenter of a triangle can be calculated by the equation:

$Distance = \sqrt{R^2 - 2rR}$

Below is the implementation of the above approach:

C++14

// C++14 program for the above approach
#include <bits/stdc++.h> 
using namespace std;
 
// Function returns the required distance
double distance(int r, int R) 
{ 
    double d = sqrt(pow(R, 2) - 
                       (2 * r * R)); 
                             
    return d; 
} 
 
// Driver code 
int main() 
{ 
     
    // Length of Inradius 
    int r = 2; 
     
    // Length of Circumradius 
    int R = 5; 
 
    cout << (round(distance(r, R) * 100.0) / 100.0); 
} 
 
// This code is contributed by sanjoy_62

Java

// Java program for the above approach 
import java.util.*;
 
class GFG{
     
// Function returns the required distance
static double distance(int r,int R)
{
    double d = Math.sqrt(Math.pow(R, 2) -
                         (2 * r * R));
                          
    return d;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Length of Inradius
    int r = 2;
     
    // Length of Circumradius
    int R = 5;
 
    System.out.println(Math.round(
        distance(r, R) * 100.0) / 100.0);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
import math
 
# Function returns the required distance
def distance(r,R):
 
    d = math.sqrt( (R**2) - (2 * r * R))
     
    return d
 
# Driver Code
 
# Length of Inradius
r = 2
 
# Length of Circumradius
R = 5
 
print(round(distance(r,R),2))

C#

// C# program for the above approach 
using System;
 
class GFG{
     
// Function returns the required distance
static double distance(int r, int R)
{
    double d = Math.Sqrt(Math.Pow(R, 2) -
                         (2 * r * R));
                         
    return d;
}
 
// Driver code
public static void Main(string[] args)
{
     
    // Length of Inradius
    int r = 2;
     
    // Length of Circumradius
    int R = 5;
     
    Console.Write(Math.Round(
        distance(r, R) * 100.0) / 100.0);
}
}
 
// This code is contributed by rutvik_56

Javascript

<script>
 
// Javascript program for
// the above approach
 
// Function returns the required distance
function distance(r, R)
{
    let d = Math.sqrt(Math.pow(R, 2) -
                         (2 * r * R));
                            
    return d;
}
 
// Driver code
 
    // Length of Inradius
    let r = 2;
       
    // Length of Circumradius
    let R = 5;
   
    document.write(Math.round(
        distance(r, R) * 100.0) / 100.0);
      
     // This code is contributed by susmitakundugoaldanga.
</script>

Output:

2.24

Time Complexity: O(logn) since time complexity of sqrt is O(logn)
Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!