Area of square Circumscribed by Circle

Given the radius(r) of circle then find the area of square which is Circumscribed by circle.
Examples: 
 

Input : r = 3
Output :Area of square = 18

Input :r = 6
Output :Area of square = 72

All four sides of a square are of equal length and all four angles are 90 degree. The circle is circumscribed on a given square shown by a shaded region in the below diagram. 
 

Properties of Circumscribed circle are as follows: 
 

• The center of the circumcircle is the point where the two diagonals of a square meet.
• Circumscribed circle of a square is made through the four vertices of a square.
• The radius of a circumcircle of a square is equal to the radius of a square.

Formula used to calculate the area of circumscribed square is: 
2 * r² 
where, r is the radius of the circle in which a square is circumscribed by circle.
How does this formula work?
Assume diagonal of square is d and length of side is a.
We know from the Pythagoras Theorem, the diagonal of a 
square is (2) times the length of a side. 
i.e d² = a² + a² 
d = 2 * a² 
d = (2) * a 
Now, 
a = d / 2
and We know diagonal of square that are Circumscribed by 
Circle is equal to Diameter of circle. 
so Area of square = a * a 
= d / (2) * d / (2) 
= d²/ 2 
= ( 2 * r )²/ 2 ( We know d = 2 * r ) 
= 2 * r²

// C++ program to find Area of
// square Circumscribed by Circle
#include <iostream>
using namespace std;

// Function to find area of square
int find_Area(int r)
{
    return (2 * r * r);
}

// Driver code
int main() 
{
    // Radius of a circle
    int r = 3;
    
    // Call Function to find 
    // an area of square
    cout << " Area of square = " 
                 << find_Area(r);
    
    return 0;
}

// Java program to find Area of
// square Circumscribed by Circle
class GFG {
    
    // Function to find area of square
    static int find_Area(int r)
    {
        return (2 * r * r);
    }
    
    // Driver code
    public static void main(String[] args)
    {
        // Radius of a circle
        int r = 3;

        // Call Function to find
        // an area of square
        System.out.print(" Area of square = "
                             + find_Area(r));
    }
}

// This code is contributed by Anant Agarwal.

# Python program to
# find Area of
# square Circumscribed
# by Circle

# Function to find
# area of square
def find_Area(r):

    return (2 * r * r)
    
# driver code
# Radius of a circle
r = 3
    
# Call Function to find 
# an area of square
print(" Area of square = ", find_Area(r))

# This code is contributed
# by Anant Agarwal.

// C# program to find Area of
// square Circumscribed by Circle
using System;

class GFG {
    
    // Function to find area of square
    static int find_Area(int r)
    {
        return (2 * r * r);
    }
    
    // Driver code
    public static void Main()
    {
        // Radius of a circle
        int r = 3;

        // Call Function to find
        // an area of square
        Console.WriteLine(" Area of square = "
                            + find_Area(r));
    }
}

// This code is contributed by vt_m.

<script>
// Javascript program to find Area of 
// square Circumscribed by Circle 

// Function to find area of square 
function find_Area(r) 
{ 
    return (2 * r * r); 
} 

// Driver code 
 
    // Radius of a circle 
    let r = 3; 
    
    // Call Function to find 
    // an area of square 
    document.write(" Area of square = "
                + find_Area(r)); 
    
// This code is contributed by Mayank Tyagi
</script>

<?php
// PHP program to find Area of
// square Circumscribed by Circle

// Function to find area of square
function find_Area( $r)
{
    return (2 * $r * $r);
}

    // Driver code
    // Radius of a circle
    $r = 3;
    
    // Call Function to find 
    // an area of square
    echo ("Area of square = ");
    echo(find_Area($r));
    
// This code is contributed by vt_m.
?>

Output:
 

Area of square = 18

Time Complexity: O(1) 
Auxiliary Space: O(1)
Please suggest if someone has a better solution which is more efficient in terms of space and time.