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

• Difficulty Level : Medium
• Last Updated : 19 Oct, 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 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 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 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 approach using 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(1)
Auxiliary Space: O(1)

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