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

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 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 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 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# 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

<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>

2.24

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