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Number of smaller circles that can be inscribed in a larger circle

  • Last Updated : 22 Apr, 2021

Given two positive integers R1 and R2, where R1 and R2 represent the radius of the larger and smaller circles respectively, the task is to find out the number of smaller circles that can be placed inside the larger circle such that the smaller circle touches the boundary of the larger circle.

Examples:

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Input: R1 = 3, R2 = 1
Output: 6
Explanation: The radii of the circles are 3 and 1. Therefore, the circle of radius 1 can be inscribed in the circle of radius 3.



From the above representation, the total number of smaller circles that can be inscribed with touching of the boundary of the larger circle is 6.

Input: R1 = 5, R2 = 4
Output: 1

 

Approach: The given problem can be solved by finding the angle that the smaller circle of radius R2 makes at the centre of the circle with radius R1 and then divide it by 360 degrees
Follow the steps below to solve the given problem:

  • If the value of R1 is less than R2, then it is impossible to inscribe a single circle. Therefore, print 0.
  • If the value of R1 is less than 2 * R2, i.e. if the diameter of the smaller circle is greater than the radius of the larger circle, then only one circle can be inscribed. Therefore, print 1.
  • Otherwise, find the angle made by the smaller circle at the centreof the larger circle using the below formula and then, divide by 360 degrees to get the total number of circles that can be inscribed and print that value.

\frac{1}{\pi}{|sin^{-1}(\frac{R2}{R1-R2}) * 180|}

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count number of smaller
// circles that can be inscribed in
// the larger circle touching its boundary
int countInscribed(int R1, int R2)
{
    // If R2 is greater than R1
    if (R2 > R1)
        return 0;
 
    // Stores the angle made
    // by the smaller circle
    double angle;
 
    // Stores the ratio
    // of R2 / (R1 - R2)
    double ratio;
 
    // Stores the count of smaller
    // circles that can be inscribed
    int number_of_circles = 0;
 
    // Stores the ratio
    ratio = R2 / (double)(R1 - R2);
 
    // If the diameter of smaller
    // circle is greater than the
    // radius of the larger circle
    if (R1 < 2 * R2) {
        number_of_circles = 1;
    }
 
    // Otherwise
    else {
 
        // Find the angle using formula
        angle = abs(asin(ratio) * 180)
                / 3.14159265;
 
        // Divide 360 with angle
        // and take the floor value
        number_of_circles = 360
                            / (2
                               * floor(angle));
    }
 
    // Return the final result
    return number_of_circles;
}
 
// Driver Code
int main()
{
    int R1 = 3;
    int R2 = 1;
 
    cout << countInscribed(R1, R2);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
  
// Function to count number of smaller
// circles that can be inscribed in
// the larger circle touching its boundary
static int countInscribed(int R1, int R2)
{
     
    // If R2 is greater than R1
    if (R2 > R1)
        return 0;
 
    // Stores the angle made
    // by the smaller circle
    double angle;
 
    // Stores the ratio
    // of R2 / (R1 - R2)
    double ratio;
 
    // Stores the count of smaller
    // circles that can be inscribed
    int number_of_circles = 0;
 
    // Stores the ratio
    ratio = R2 / (double)(R1 - R2);
 
    // If the diameter of smaller
    // circle is greater than the
    // radius of the larger circle
    if (R1 < 2 * R2)
    {
        number_of_circles = 1;
    }
 
    // Otherwise
    else
    {
         
        // Find the angle using formula
        angle = Math.abs(Math.asin(ratio) * 180) /
                3.14159265;
 
        // Divide 360 with angle
        // and take the floor value
        number_of_circles = (int)(360 /
        (2 * Math.floor(angle)));
    }
     
    // Return the final result
    return number_of_circles;
}
 
// Driver Code
public static void main(String args[])
{
    int R1 = 3;
    int R2 = 1;
     
    System.out.println(countInscribed(R1, R2));
}
}
 
// This code is contributed by ipg2016107

Python3




# Python3 program for the above approach
import math
 
# Function to count number of smaller
# circles that can be inscribed in
# the larger circle touching its boundary
def countInscribed(R1, R2):
     
    # If R2 is greater than R1
    if (R2 > R1):
        return 0
 
    # Stores the angle made
    # by the smaller circle
    angle = 0
 
    # Stores the ratio
    # of R2 / (R1 - R2)
    ratio = 0
 
    # Stores the count of smaller
    # circles that can be inscribed
    number_of_circles = 0
 
    # Stores the ratio
    ratio = R2 / (R1 - R2)
 
    # If the diameter of smaller
    # circle is greater than the
    # radius of the larger circle
    if (R1 < 2 * R2):
        number_of_circles = 1
     
    # Otherwise
    else:
 
        # Find the angle using formula
        angle = (abs(math.asin(ratio) * 180) /
                 3.14159265)
        
        # Divide 360 with angle
        # and take the floor value
        number_of_circles = (360 / (2 *
               math.floor(angle)))
     
    # Return the final result
    return number_of_circles
 
# Driver Code
if __name__ == "__main__":
 
    R1 = 3
    R2 = 1
 
    print (int(countInscribed(R1, R2)))
 
# This code is contributed by ukasp

C#




// C# program for the above approach
using System;
 
class GFG{
  
// Function to count number of smaller
// circles that can be inscribed in
// the larger circle touching its boundary
static int countInscribed(int R1, int R2)
{
     
    // If R2 is greater than R1
    if (R2 > R1)
        return 0;
 
    // Stores the angle made
    // by the smaller circle
    double angle;
 
    // Stores the ratio
    // of R2 / (R1 - R2)
    double ratio;
 
    // Stores the count of smaller
    // circles that can be inscribed
    int number_of_circles = 0;
 
    // Stores the ratio
    ratio = R2 / (double)(R1 - R2);
 
    // If the diameter of smaller
    // circle is greater than the
    // radius of the larger circle
    if (R1 < 2 * R2)
    {
        number_of_circles = 1;
    }
 
    // Otherwise
    else
    {
         
        // Find the angle using formula
        angle = Math.Abs(Math.Asin(ratio) * 180) /
                3.14159265;
 
        // Divide 360 with angle
        // and take the floor value
        number_of_circles = (int)(360 /
        (2 * Math.Floor(angle)));
    }
     
    // Return the final result
    return number_of_circles;
}
 
// Driver Code
public static void Main()
{
    int R1 = 3;
    int R2 = 1;
     
    Console.WriteLine(countInscribed(R1, R2));
}
}
 
// This code is contributed by mohit kumar 29

Javascript




<script>
 
        // Javascript program for the above approach
 
        // Function to count number of smaller
        // circles that can be inscribed in
        // the larger circle touching its boundary
        function countInscribed(R1, R2)
        {
            // If R2 is greater than R1
            if (R2 > R1)
                return 0;
 
            // Stores the angle made
            // by the smaller circle
            let angle;
 
            // Stores the ratio
            // of R2 / (R1 - R2)
            let ratio;
 
            // Stores the count of smaller
            // circles that can be inscribed
            let number_of_circles = 0;
 
            // Stores the ratio
            ratio = R2 / (R1 - R2);
 
            // If the diameter of smaller
            // circle is greater than the
            // radius of the larger circle
            if (R1 < 2 * R2) {
                number_of_circles = 1;
            }
 
            // Otherwise
            else {
 
                // Find the angle using formula
                angle = Math.abs(Math.asin(ratio) * 180)
                    / 3.14159265;
 
                // Divide 360 with angle
                // and take the floor value
                number_of_circles = 360
                    / (2
                        * Math.floor(angle));
            }
 
            // Return the final result
            return number_of_circles;
        }
 
        // Driver Code
 
        let R1 = 3;
        let R2 = 1;
 
        document.write(countInscribed(R1, R2))
 
        // This code is contributed by Hritik
         
    </script>
Output: 
6

 

Time Complexity: O(1)
Auxiliary Space: O(1)




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