# Radius of the inscribed circle within three tangent circles

• Difficulty Level : Medium
• Last Updated : 07 Aug, 2022

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

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 formuladouble 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 circledouble 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 codeint 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 approachimport 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

## Python3

 # Python3 implementation of the approachimport math # Radius of the 3 given circlesr1 = 0r2 = 0r3 = 0 # Calculation of area of a# triangle by Heron's formuladef area(a, b, c):         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 circledef binary_search():         global r1, r2, r3     # Area of main triangle    s = area(r1 + r2, r2 + r3, r3 + r1)    l = 0    h = s / (r1 + r2 + r3)     # Loop runs until l and h    # becomes approximately equal    while (h - l > 0.00000001):        mid = (l + h) / 2         # Area of smaller triangles        s1 = area(mid + r1, mid + r2, r1 + r2)        s2 = area(mid + r1, mid + r3, r1 + r3)        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 # Taking r1, r2, r3 as inputr1 = 1r2 = 2r3 = 3 # Call to function binary searchprint("{0:.6f}".format(binary_search())) # This code is contributed by avanitrachhadiya2155

## C#

 // C# implementation of the approachusing 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

## Javascript

 

Output:

0.260870

Time Complexity: O(logn)

Auxiliary Space: O(1)

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.

## C++

 // C++ implementation of the approach#include using namespace std; // Driver codeint 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 approachclass 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

## Python3

 # Python3 implementation of the approachfrom math import sqrt # Driver code # Radius of the 3 given circles declared as double. # Taking r1, r2, r3 as inputr1 = 1r2 = 2r3 = 3 # Calculation of r4 using formula given abover4 = (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 approachusing 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

## Javascript

 

Output:

0.260870

Time Complexity: O(log(n)) as inbuilt sqrt function is being used which has time complexity of log(n)

Auxiliary Space: O(1)

Reference :

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