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:**

Input:R1 = 3, R2 = 1Output:6Explanation: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 = 4Output: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.

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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