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

• Difficulty Level : Easy
• Last Updated : 09 Sep, 2022

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:

Below is the implementation of the above approach:

## C++14

 // C++14 program for the above approach#include using namespace std; // Function returns the required distancedouble distance(int r, int R){    double d = sqrt(pow(R, 2) -                       (2 * r * R));                                 return d;} // Driver codeint 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 approachimport java.util.*; class GFG{     // Function returns the required distancestatic double distance(int r,int R){    double d = Math.sqrt(Math.pow(R, 2) -                         (2 * r * R));                              return d;} // Driver codepublic 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 approachimport math # Function returns the required distancedef distance(r,R):     d = math.sqrt( (R**2) - (2 * r * R))         return d # Driver Code # Length of Inradiusr = 2 # Length of CircumradiusR = 5 print(round(distance(r,R),2))

## C#

 // C# program for the above approachusing System; class GFG{     // Function returns the required distancestatic double distance(int r, int R){    double d = Math.Sqrt(Math.Pow(R, 2) -                         (2 * r * R));                             return d;} // Driver codepublic 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

 

Output:

2.24

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

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