Find area of the larger circle when radius of the smaller circle and difference in the area is given

Given two integers r and d where r is the radius of the smaller circle and d is the difference of the area of this circle with some larger radius circle. The task is to find the area of the larger circle.
Examples:

Input: r = 4, d = 5 
Output: 55.24 
Area of the smaller circle = 3.14 * 4 * 4 = 50.24 
55.24 – 50.24 = 5
Input: r = 12, d = 3 
Output: 455.16 

Approach: Let radius of the smaller and the larger circles be r and R respectively and the difference in the areas is given to be d i.e. PI * R² – PI * r² = d where PI = 3.14
Or, R² = (d / PI) + r². 
Now, area of the bigger circle can be calculated as PI * R².
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

const double PI = 3.14;

// Function to return the area
// of the bigger circle
double find_area(int r, int d)
{
    // Find the radius of
    // the bigger circle
    double R = d / PI;
    R += pow(r, 2);
    R = sqrt(R);

    // Calculate the area of
    // the bigger circle
    double area = PI * pow(R, 2);
    return area;
}

// Driver code
int main()
{
    int r = 4, d = 5;

    cout << find_area(r, d);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{
    static double PI = 3.14;

    // Function to return the area
    // of the bigger circle
    static double find_area(int r, int d)
    {
        // Find the radius of
        // the bigger circle
        double R = d / PI;
        R += Math.pow(r, 2);
        R = Math.sqrt(R);

        // Calculate the area of
        // the bigger circle
        double area = PI * Math.pow(R, 2);
        return area;
    }

    // Driver code
    public static void main(String[] args)
    {
        int r = 4, d = 5;

        System.out.println(find_area(r, d));
    }
}

// This code is contributed by PrinciRaj1992

Python3

# Python 3 implementation of the approach
PI = 3.14
from math import pow, sqrt

# Function to return the area
# of the bigger circle
def find_area(r, d):

    # Find the radius of
    # the bigger circle
    R = d / PI
    R += pow(r, 2)
    R = sqrt(R)

    # Calculate the area of
    # the bigger circle
    area = PI * pow(R, 2)
    return area

# Driver code
if __name__ == '__main__':
    r = 4
    d = 5

    print(find_area(r, d))

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;

public class GFG
{
    static double PI = 3.14;

    // Function to return the area
    // of the bigger circle
    static double find_area(int r, int d)
    {
        // Find the radius of
        // the bigger circle
        double R = d / PI;
        R += Math.Pow(r, 2);
        R = Math.Sqrt(R);

        // Calculate the area of
        // the bigger circle
        double area = PI * Math.Pow(R, 2);
        return area;
    }

    // Driver code
    static public void Main ()
    {

        int r = 4, d = 5;
        Console.Write(find_area(r, d));
    }
}

// This code is contributed by ajit.

PHP

<?php
// PHP implementation of the approach
const PI = 3.14;

// Function to return the area
// of the bigger circle
function find_area($r, $d)
{

    // Find the radius of
    // the bigger circle
    $R = $d / PI;
    $R += pow($r, 2);
    $R = sqrt($R);

    // Calculate the area of
    // the bigger circle
    $area = PI * pow($R, 2);
    return $area;
}

// Driver Code
$r = 4;
$d = 5;

echo (find_area($r, $d));

// This code is contributed by Naman_Garg
?>

Javascript

<script>

// Javascript implementation of the approach

 let PI = 3.14;

// Function to return the area
// of the bigger circle
function find_area(r, d)
{
    // Find the radius of
    // the bigger circle
    let R = d / PI;
    R += Math.pow(r, 2);
    R = Math.sqrt(R);

    // Calculate the area of
    // the bigger circle
    let area = PI * Math.pow(R, 2);
    return area;
}

// Driver code
let r = 4, d = 5;
    document.write( find_area(r,d).toFixed(2));

// This code contributed by Rajput-Ji

</script>

Output:

55.24

Time complexity: O(logR)

Auxiliary Space: O(1)