# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 `  `

Output:

```55.24
```

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