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

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 <bits/stdc++.h>
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

<script>

// Javascript program of the above approach

 // Function to calculate radius of the circle
// having area equal to sum of the area of
// two circles with given radii
function findRadius(r1, r2)
{
    let 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

    // Given radius
    let r1 = 8, r2 = 6;

    // Prints the radius of
    // the required circle
    document.write(findRadius(r1, r2));

</script>

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.