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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?> 

chevron_right


Output:

55.24


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.