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Program to calculate area of inner circle which passes through center of outer circle and touches its circumference

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  • Difficulty Level : Hard
  • Last Updated : 18 Jun, 2022

Given a circle C1 and it’s a radius r1. And one another circle C2 whose passes through center of circle C1 and touch the circumference of circle C1. The task is to find out the area of circle C2
Examples: 
 

Input: r1 = 4
Output:Area of circle c2 = 12.56

Input: r1 = 7
Output:Area of circle c2 = 38.465

 

Approach: 
Radius r2 of circle C2 is \ r2 = r1/2 \
So we know that the area of circle is \ Area = \pi r^2 \     .
Below is the implementation of the above approach:
 

C++




// C++ implementation of the above approach
#include<bits/stdc++.h>
#include <iostream>
using namespace std;
 
// Function calculate the area of the inner circle
double innerCirclearea(double radius)
{
 
    // the radius cannot be negative
    if (radius < 0)
    {
        return -1;
    }
 
    // area of the circle
    double r = radius / 2;
    double Area = (3.14 * pow(r, 2));
 
    return Area;
}
 
// Driver Code
int main()
{
     
    double radius = 4;
    cout << ("Area of circle c2 = ",
                innerCirclearea(radius));
    return 0;
}
 
// This code is contributed by jit_t.

Java




// Java implementation of the above approach
 
class GFG {
 
    // Function calculate the area of the inner circle
    static double innerCirclearea(double radius)
    {
 
        // the radius cannot be negative
        if (radius < 0) {
            return -1;
        }
 
        // area of the circle
        double r = radius / 2;
        double Area = (3.14 * Math.pow(r, 2));
 
        return Area;
    }
 
    // Driver Code
    public static void main(String arr[])
    {
        double radius = 4;
        System.out.println("Area of circle c2 = "
                           + innerCirclearea(radius));
    }
}

Python3




# Python3 implementation of the above approach
 
# Function calculate the area of the inner circle
def innerCirclearea(radius) :
 
    # the radius cannot be negative
    if (radius < 0) :
        return -1;
         
    # area of the circle
    r = radius / 2;
    Area = (3.14 * pow(r, 2));
 
    return Area;
     
# Driver Code
if __name__ == "__main__" :
     
    radius = 4;
    print("Area of circle c2 =",
           innerCirclearea(radius));
 
# This code is contributed by AnkitRai01

C#




// C# Implementation of the above approach
using System;
     
class GFG
{
 
    // Function calculate the area
    // of the inner circle
    static double innerCirclearea(double radius)
    {
 
        // the radius cannot be negative
        if (radius < 0)
        {
            return -1;
        }
 
        // area of the circle
        double r = radius / 2;
        double Area = (3.14 * Math.Pow(r, 2));
 
        return Area;
    }
 
    // Driver Code
    public static void Main(String []arr)
    {
        double radius = 4;
        Console.WriteLine("Area of circle c2 = " +
                          innerCirclearea(radius));
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// JavaScript implementation of the above approach
 
// Function calculate the area of the inner circle
function innerCirclearea(radius)
{
 
    // the radius cannot be negative
    if (radius < 0)
    {
        return -1;
    }
 
    // area of the circle
    let r = radius / 2;
    let Area = (3.14 * Math.pow(r, 2));
 
    return Area;
}
 
// Driver Code
 
     
    let radius = 4;
    document.write("Area of circle c2 = " +
                innerCirclearea(radius));
 
// This code is contributed by Surbhi Tyagi.
 
</script>

Output: 

Area of circle c2 = 12.56

 

Time Complexity : O(log r) ,where r is the radius of circle.

Auxiliary Space : O(1) ,as we are not using any extra space.


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