# Radius of the inscribed circle within three tangent circles

There are 4 circles with positive integer radius r1, r2, r3 and r4 as shown in the figure below. The task is to find the radius r4 of the circle formed by three circles when radius r1, r2, r3 are given.
(Note that the circles in the picture above are tangent to each other.)

Examples:

Input: r1 = 1, r2 = 1, r3 = 1
Output: 0.154701

Input: r1 = 23, r2 = 46, r3 = 69
Output: 6.000000

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach 1: (Using Binary Search) :

1. The policy is to join the centers of all the circles and form 4 triangles
2. After the triangles are formed, equate the sum of areas of the three smaller triangles with the main triangle as far as possible using binary search.

Analysis of the mentioned approach:

1. This method works because initially there are 4 triangles as pointed in the above image.
2. The main triangle with sides and the three smaller triangles with sides .
3. The main triangle consists of the small ones so the area of the main triangle is the sum of the areas of the smaller ones.

Forming a search space:
Here binary search. The value for r can be chosen and the sum of the areas of all three smaller triangles can be computed and compared with the area of the main triangle.

1. Choosing lower bound 2. Choosing upper bound By intuition, the upper bound value for r4 as the radius of the inscribed circle into the triangle is less than:
rupper_bound Now Binary Search can be applied at the following search space.

Below is the implementation of the problem using the above approach.

## C++

 // C++ implementation of the approach  #include  using namespace std;     // Radius of the 3 given circles  // declared as double.  double r1, r2, r3;     // Calculation of area of a triangle by Heron's formula  double area(double a, double b, double c)  {      double p = (a + b + c) / 2;      return sqrt(p) * sqrt(p - a) * sqrt(p - b) * sqrt(p - c);  }     // Applying binary search to find the  // radius r4 of the required circle  double binary_search()  {      // Area of main triangle      double s = area(r1 + r2, r2 + r3, r3 + r1);      double l = 0, h = s / (r1 + r2 + r3);      // Loop runs until l and h becomes approximately equal      while (h - l >= 1.e-7) {          double mid = (l + h) / 2;             // Area of smaller triangles          double s1 = area(mid + r1, mid + r2, r1 + r2);          double s2 = area(mid + r1, mid + r3, r1 + r3);          double s3 = area(mid + r2, mid + r3, r2 + r3);             // If sum of smaller triangles          // is less than main triangle          if (s1 + s2 + s3 < s) {              l = mid;          }          // If sum of smaller triangles is          // greater than or equal to main triangle          else {              h = mid;          }      }      return (l + h) / 2;  }  // Driver code  int main()  {      // Taking r1, r2, r3 as input      r1 = 1.0;      r2 = 2.0;      r3 = 3.0;      // Call to function binary search      cout << fixed << setprecision(6) << binary_search() << endl;      return 0;  }

## Java

 // Java implementation of the approach  import java.util.*;     class GFG  {         // Radius of the 3 given circles      // declared as double.      static double r1, r2, r3;         // Calculation of area of a triangle by Heron's formula      static double area(double a, double b, double c)       {          double p = (a + b + c) / 2;          return Math.sqrt(p) * Math.sqrt(p - a) *                  Math.sqrt(p - b) * Math.sqrt(p - c);      }         // Applying binary search to find the      // radius r4 of the required circle      static double binary_search()      {          // Area of main triangle          double s = area(r1 + r2, r2 + r3, r3 + r1);          double l = 0, h = s / (r1 + r2 + r3);                     // Loop runs until l and h becomes approximately equal          while (h - l >= 1.e-7)          {              double mid = (l + h) / 2;                 // Area of smaller triangles              double s1 = area(mid + r1, mid + r2, r1 + r2);              double s2 = area(mid + r1, mid + r3, r1 + r3);              double s3 = area(mid + r2, mid + r3, r2 + r3);                 // If sum of smaller triangles              // is less than main triangle              if (s1 + s2 + s3 < s)               {                  l = mid;              }                             // If sum of smaller triangles is              // greater than or equal to main triangle              else             {                  h = mid;              }          }          return (l + h) / 2;      }         // Driver code      public static void main(String[] args)      {          // Taking r1, r2, r3 as input          r1 = 1.0;          r2 = 2.0;          r3 = 3.0;                     // Call to function binary search          System.out.printf("%.6f", binary_search());      }  }     // This code is contributed by 29AjayKumar

## C#

 // C# implementation of the approach  using System;     class GFG  {         // Radius of the 3 given circles      // declared as double.      static double r1, r2, r3;         // Calculation of area of a triangle by Heron's formula      static double area(double a, double b, double c)       {          double p = (a + b + c) / 2;          return Math.Sqrt(p) * Math.Sqrt(p - a) *               Math.Sqrt(p - b) * Math.Sqrt(p - c);      }         // Applying binary search to find the      // radius r4 of the required circle      static double binary_search()      {          // Area of main triangle          double s = area(r1 + r2, r2 + r3, r3 + r1);          double l = 0, h = s / (r1 + r2 + r3);                     // Loop runs until l and h           // becomes approximately equal          while (h - l > 0.00000001)          {              double mid = (l + h) / 2;                 // Area of smaller triangles              double s1 = area(mid + r1, mid + r2, r1 + r2);              double s2 = area(mid + r1, mid + r3, r1 + r3);              double s3 = area(mid + r2, mid + r3, r2 + r3);                 // If sum of smaller triangles              // is less than main triangle              if (s1 + s2 + s3 < s)               {                  l = mid;              }                             // If sum of smaller triangles is              // greater than or equal to main triangle              else             {                  h = mid;              }          }          return (l + h) / 2;      }         // Driver code      public static void Main(String[] args)      {          // Taking r1, r2, r3 as input          r1 = 1.0;          r2 = 2.0;          r3 = 3.0;                     // Call to function binary search          Console.Write("{0:F6}", binary_search());      }  }     // This code is contributed by Rajput-Ji

Output:

0.260870


Approach 2: (Using Descartes’ Theorem)

1. According to Descartes’ Theorem, the reciprocals of radii, or “curvatures”, of these circles satisfy the following relation. 2. If are known, one can solve for, 3. On solving the above equation; Below is the implementation of the problem using the above formula.

## CPP

 // C++ implementation of the approach  #include  using namespace std;     // Driver code  int main()  {      // Radius of the 3 given circles declared as double.      double r1, r2, r3;         // Taking r1, r2, r3 as input      r1 = 1;      r2 = 2;      r3 = 3;         // Calculation of r4 using formula given above      double r4 = (r1 * r2 * r3)                  / (r1 * r2 + r2 * r3 + r1 * r3                     + 2.0 * sqrt(r1 * r2 * r3 * (r1 + r2 + r3)));      cout << fixed << setprecision(6) << r4 << '\n';      return 0;  }

## Java

 // Java implementation of the approach  class GFG   {         // Driver code      public static void main(String[] args)       {          // Radius of the 3 given circles declared as double.          double r1, r2, r3;             // Taking r1, r2, r3 as input          r1 = 1;          r2 = 2;          r3 = 3;             // Calculation of r4 using formula given above          double r4 = (r1 * r2 * r3) /                       (r1 * r2 + r2 * r3 + r1 * r3 + 2.0 *                       Math.sqrt(r1 * r2 * r3 * (r1 + r2 + r3)));          System.out.printf("%.6f", r4);      }  }     // This code is contributed by 29AjayKumar

## Python

 # Python3 implementation of the approach  from math import sqrt     # Driver code     # Radius of the 3 given circles declared as double.     # Taking r1, r2, r3 as input  r1 = 1 r2 = 2 r3 = 3    # Calculation of r4 using formula given above  r4 = (r1 * r2 * r3)/ (r1 * r2 + r2 * r3 + r1 * r3              + 2.0 * sqrt(r1 * r2 * r3 * (r1 + r2 + r3)))  print(round(r4, 6))     # This code is contributed by mohit kumar 29

## C#

 // C# implementation of the approach  using System;     class GFG   {          // Driver code      public static void Main(String[] args)       {          // Radius of the 3 given circles declared as double.          double r1, r2, r3;              // Taking r1, r2, r3 as input          r1 = 1;          r2 = 2;          r3 = 3;              // Calculation of r4 using formula given above          double r4 = (r1 * r2 * r3) /                       (r1 * r2 + r2 * r3 + r1 * r3 + 2.0 *                       Math.Sqrt(r1 * r2 * r3 * (r1 + r2 + r3)));          Console.Write("{0:F6}", r4);      }  }     // This code contributed by PrinciRaj1992

Output:

0.260870


Reference :

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.