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

• Last Updated : 16 Mar, 2021

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 ``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

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## Javascript

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Output:
`55.24`

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