Given a circle of radius r and center in point(x1, y1) and given a point(x2, y2). The task is move center of circle from given center (x1, y1) to target (x2, y2) using minimum number of steps. In one step, we can put a pin on the border of the circle at any point, then rotate the circle around that pin by any angle and finally remove the pin. **Examples :**

Input : r = 2 x1 = 0, y1 = 0 x2 = 0, y2 = 4 Output :1 Input : r = 1 x1 = 1, y1 = 1, x2 = 4, y2 = 4 Output : 3

Let’s consider a straight line between the two centers. Clearly to move the center with maximum distance we need to rotate it around the line and with 180 degrees. So the maximum distance we can move the center each time is 2 * r. Let’s continue moving the center with 2 * r distance each time until the two circles intersects. Now obviously we can make the center moves into its final position by rotating it around one of the intersection points of the two circles until it reaches the final destination.

Every time we make the circle moves 2 * r times except the last move it’ll be < 2 * r. Let the initial distance between the two points be d. Solution to the problem will be ceil(d/2*r).

## C++

`// C++ program to find minimum number of` `// revolutions to reach a target center` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Minimum revolutions to move center from` `// (x1, y1) to (x2, y2)` `int` `minRevolutions(` `double` `r, ` `int` `x1, ` `int` `y1,` ` ` `int` `x2, ` `int` `y2)` `{` ` ` `double` `d = ` `sqrt` `((x1 - x2)*(x1 - x2) +` ` ` `(y1 - y2)*(y1 - y2));` ` ` `return` `ceil` `(d/(2*r));` `}` `// Driver code` `int` `main()` `{` ` ` `int` `r = 2, x1 = 0, y1 = 0, x2 = 0, y2 = 4;` ` ` `cout << minRevolutions(r, x1, y1, x2, y2);` ` ` `return` `0;` `}` |

## Java

`// Java program to find minimum number of` `// revolutions to reach a target center` `class` `GFG {` ` ` ` ` `// Minimum revolutions to move center` ` ` `// from (x1, y1) to (x2, y2)` ` ` `static` `double` `minRevolutions(` `double` `r,` ` ` `int` `x1, ` `int` `y1, ` `int` `x2, ` `int` `y2)` ` ` `{` ` ` ` ` `double` `d = Math.sqrt((x1 - x2)` ` ` `* (x1 - x2) + (y1 - y2)` ` ` `* (y1 - y2));` ` ` ` ` `return` `Math.ceil(d / (` `2` `* r));` ` ` `}` ` ` ` ` `// Driver Program to test above function` ` ` `public` `static` `void` `main(String arg[]) {` ` ` ` ` `int` `r = ` `2` `, x1 = ` `0` `, y1 = ` `0` `;` ` ` `int` `x2 = ` `0` `, y2 = ` `4` `;` ` ` ` ` `System.out.print((` `int` `)minRevolutions(r,` ` ` `x1, y1, x2, y2));` ` ` `}` `}` `// This code is contributed by Anant Agarwal.` |

## Python3

`# Python program to find` `# minimum number of` `# revolutions to reach` `# a target center` `import` `math` `# Minimum revolutions to move center from` `# (x1, y1) to (x2, y2)` `def` `minRevolutions(r,x1,y1,x2,y2):` ` ` `d ` `=` `math.sqrt((x1 ` `-` `x2)` `*` `(x1 ` `-` `x2) ` `+` ` ` `(y1 ` `-` `y2)` `*` `(y1 ` `-` `y2))` ` ` `return` `math.ceil(d` `/` `(` `2` `*` `r))` `# Driver code` `r ` `=` `2` `x1 ` `=` `0` `y1 ` `=` `0` `x2 ` `=` `0` `y2 ` `=` `4` `print` `(minRevolutions(r, x1, y1, x2, y2))` `# This code is contributed` `# by Anant Agarwal.` |

## C#

`// C# program to find minimum number of` `// revolutions to reach a target center` `using` `System;` `class` `GFG {` ` ` ` ` `// Minimum revolutions to move center` ` ` `// from (x1, y1) to (x2, y2)` ` ` `static` `double` `minRevolutions(` `double` `r,` ` ` `int` `x1, ` `int` `y1, ` `int` `x2, ` `int` `y2)` ` ` `{` ` ` ` ` `double` `d = Math.Sqrt((x1 - x2)` ` ` `* (x1 - x2) + (y1 - y2)` ` ` `* (y1 - y2));` ` ` ` ` `return` `Math.Ceiling(d / (2 * r));` ` ` `}` ` ` ` ` `// Driver Program to test above function` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `r = 2, x1 = 0, y1 = 0;` ` ` `int` `x2 = 0, y2 = 4;` ` ` ` ` `Console.Write((` `int` `)minRevolutions(r,` ` ` `x1, y1, x2, y2));` ` ` `}` `}` `// This code is contributed by nitin mittal.` |

## PHP

`<?php` `// PHP program to find minimum` `// number of revolutions to reach` `// a target center` `// Minimum revolutions to move` `// center from (x1, y1) to (x2, y2)` `function` `minRevolutions(` `$r` `, ` `$x1` `, ` `$y1` `,` ` ` `$x2` `, ` `$y2` `)` `{` ` ` `$d` `= sqrt((` `$x1` `- ` `$x2` `) * (` `$x1` `- ` `$x2` `) +` ` ` `(` `$y1` `- ` `$y2` `) * (` `$y1` `- ` `$y2` `));` ` ` `return` `ceil` `(` `$d` `/ (2 * ` `$r` `));` `}` `// Driver code` `$r` `= 2; ` `$x1` `= 0; ` `$y1` `= 0; ` `$x2` `= 0; ` `$y2` `= 4;` `echo` `minRevolutions(` `$r` `, ` `$x1` `, ` `$y1` `, ` `$x2` `, ` `$y2` `);` `// This code is contributed by nitin mittal.` `?>` |

## Javascript

`<script>` `// Javascript program to find minimum number of` `// revolutions to reach a target center` `// Minimum revolutions to move center from` `// (x1, y1) to (x2, y2)` `function` `minRevolutions(r, x1, y1, x2, y2)` `{` ` ` `let d = Math.sqrt((x1 - x2)*(x1 - x2) +` ` ` `(y1 - y2)*(y1 - y2));` ` ` `return` `Math.ceil(d/(2*r));` `}` `// Driver code` ` ` `let r = 2, x1 = 0, y1 = 0, x2 = 0, y2 = 4;` ` ` `document.write(minRevolutions(r, x1, y1, x2, y2));` `</script>` |

**Output :**

1

**Time Complexity :** O(1)

This article is contributed by **Rakesh Kumar**. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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