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Distance between Incenter and Circumcenter of a triangle using Inradius and Circumradius
  • Last Updated : 19 Apr, 2021

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 incentre and circumcentre 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(1) 
Auxiliary Space: O(1)
 

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