# Radius of a circle having area equal to the sum of area of the circles having given radii

• Difficulty Level : Hard
• Last Updated : 25 Sep, 2022

Given two integers r1 and r2, representing the radius of two circles, the task is to find the radius of the circle having area equal to the sum of the area of the two circles having given radii.

Examples:

Input:
r1 = 8
r2 = 6
Output:
10
Explanation:
Area of circle with radius 8 = 201.061929
Area of a circle with radius 6 = 113.097335
Area of a circle with radius 10 = 314.159265

Input:
r1 = 2
r2 = 2
Output:
2.82843

Approach: Follow the steps below to solve the problem:

1. Calculate area of the first circle is a1 = 3.14 * r1 * r1.
2. Calculate area of the second circle is a2 = 3.14 * r2 * r2.
3. Therefore, area of the third circle is a1 + a2.
4. Radius of the third circle is sqrt(a3 / 3.14)

Below is the implementation of the following approach.

## C++14

 `// C++ implementation of``// the above approach` `#include ``using` `namespace` `std;` `// Function to calculate radius of the circle``// having area equal to sum of the area of``// two circles with given radii``double` `findRadius(``double` `r1, ``double` `r2)``{``    ``double` `a1, a2, a3, r3;` `    ``// Area of first circle``    ``a1 = 3.14 * r1 * r1;` `    ``// Area of second circle``    ``a2 = 3.14 * r2 * r2;` `    ``// Area of third circle``    ``a3 = a1 + a2;` `    ``// Radius of third circle``    ``r3 = ``sqrt``(a3 / 3.14);` `    ``return` `r3;``}` `// Driver Code``int` `main()``{``    ``// Given radius``    ``double` `r1 = 8, r2 = 6;` `    ``// Prints the radius of``    ``// the required circle``    ``cout << findRadius(r1, r2);` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG``{``  ` `// Function to calculate radius of the circle``// having area equal to sum of the area of``// two circles with given radii``static` `double` `findRadius(``double` `r1, ``double` `r2)``{``    ``double` `a1, a2, a3, r3;` `    ``// Area of first circle``    ``a1 = ``3.14` `* r1 * r1;` `    ``// Area of second circle``    ``a2 = ``3.14` `* r2 * r2;` `    ``// Area of third circle``    ``a3 = a1 + a2;` `    ``// Radius of third circle``    ``r3 = Math.sqrt(a3 / ``3.14``);``    ``return` `r3;``}``   ` `// Driver code``public` `static` `void` `main(String[] args)``{``  ` `    ``// Given radius``    ``double` `r1 = ``8``, r2 = ``6``;` `    ``// Prints the radius of``    ``// the required circle``    ``System.out.println((``int``)findRadius(r1, r2));``}``}` `// This code is contributed by code_hunt.`

## Python3

 `# Python program to implement``# the above approach` `# Function to calculate radius of the circle``# having area equal to sum of the area of``# two circles with given radii``def` `findRadius(r1, r2):``    ``a1, a2, a3, r3 ``=` `0``, ``0``, ``0``, ``0``;` `    ``# Area of first circle``    ``a1 ``=` `3.14` `*` `r1 ``*` `r1;` `    ``# Area of second circle``    ``a2 ``=` `3.14` `*` `r2 ``*` `r2;` `    ``# Area of third circle``    ``a3 ``=` `a1 ``+` `a2;` `    ``# Radius of third circle``    ``r3 ``=` `((a3 ``/` `3.14``)``*``*``(``1``/``2``));``    ``return` `r3;` `# Driver code``if` `__name__ ``=``=` `'__main__'``:` `    ``# Given radius``    ``r1 ``=` `8``; r2 ``=` `6``;` `    ``# Prints the radius of``    ``# the required circle``    ``print``(``int``(findRadius(r1, r2)));` `# This code is contributed by 29AjayKumar`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG``{` `// Function to calculate radius of the circle``// having area equal to sum of the area of``// two circles with given radii``static` `double` `findRadius(``double` `r1, ``double` `r2)``{``    ``double` `a1, a2, a3, r3;` `    ``// Area of first circle``    ``a1 = 3.14 * r1 * r1;` `    ``// Area of second circle``    ``a2 = 3.14 * r2 * r2;` `    ``// Area of third circle``    ``a3 = a1 + a2;` `    ``// Radius of third circle``    ``r3 = Math.Sqrt(a3 / 3.14);``    ``return` `r3;``}` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ` `    ``// Given radius``    ``double` `r1 = 8, r2 = 6;` `    ``// Prints the radius of``    ``// the required circle``    ``Console.WriteLine((``int``)findRadius(r1, r2));``  ``}``}` `// This code is contributed by susmitakundugoaldanga`

## Javascript

 ``

Output:

`10`

Time Complexity: O(log(a3)), time complexity of the inbuilt sqrt() function is logn.
Space Complexity: O(1) as constant space is being used.

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