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 = 5Output:55.24

Area of the smaller circle = 3.14 * 4 * 4 = 50.24

55.24 – 50.24 = 5Input:r = 12, d = 3Output: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 * R ^{2} – PI * r^{2} = d** where

**PI = 3.14**

Or,

**R**.

^{2}= (d / PI) + r^{2}Now, area of the bigger circle can be calculated as

**PI * R**.

^{2}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