Find if a point lies inside a Circle

• Difficulty Level : Easy
• Last Updated : 05 Sep, 2022

Given a circle (coordinates of centre and radius) and a point (coordinate), find if the point lies inside or on the circle, or not.

Examples :

```Input: x = 4, y = 4 // Given Point
circle_x = 1, circle_y = 1, rad = 6; // Circle
Output: Inside

Input: x = 3, y = 3 // Given Point
circle_x = 0, circle_y = 1, rad = 2; // Circle
Output: Outside```

We strongly recommend you to minimize your browser and try this yourself first.

The idea is compute distance of point from center. If distance is less than or equal to radius. the point is inside, else outside.

Below is the implementation of above idea.

C++

 `// C++ program to check if a point``// lies inside a circle or not``#include ``using` `namespace` `std;` `bool` `isInside(``int` `circle_x, ``int` `circle_y,``                   ``int` `rad, ``int` `x, ``int` `y)``{``    ``// Compare radius of circle with distance``    ``// of its center from given point``    ``if` `((x - circle_x) * (x - circle_x) +``        ``(y - circle_y) * (y - circle_y) <= rad * rad)``        ``return` `true``;``    ``else``        ``return` `false``;``}` `// Driver function``int` `main()``{``    ``int` `x = 1, y = 1;``    ``int` `circle_x = 0, circle_y = 1, rad = 2;``    ``isInside(circle_x, circle_y, rad, x, y) ?``    ``cout << ``"Inside"` `: cout << ``"Outside"``;``}`

Java

 `// Java program to check if a point lies``// inside a circle or not` `class` `GFG {` `    ``static` `boolean` `isInside(``int` `circle_x, ``int` `circle_y,``                                 ``int` `rad, ``int` `x, ``int` `y)``    ``{``        ``// Compare radius of circle with``        ``// distance of its center from``        ``// given point``        ``if` `((x - circle_x) * (x - circle_x) +``            ``(y - circle_y) * (y - circle_y) <= rad * rad)``            ``return` `true``;``        ``else``            ``return` `false``;``    ``}` `    ``// Driver Program to test above function``    ``public` `static` `void` `main(String arg[])``    ``{``        ``int` `x = ``1``, y = ``1``;``        ``int` `circle_x = ``0``, circle_y = ``1``, rad = ``2``;` `        ``if` `(isInside(circle_x, circle_y, rad, x, y))``            ``System.out.print(``"Inside"``);``        ``else``            ``System.out.print(``"Outside"``);``    ``}``}` `// This code is contributed by Anant Agarwal.`

Python3

 `# Python3 program to check if``# a point lies inside a circle``# or not` `def` `isInside(circle_x, circle_y, rad, x, y):``    ` `    ``# Compare radius of circle``    ``# with distance of its center``    ``# from given point``    ``if` `((x ``-` `circle_x) ``*` `(x ``-` `circle_x) ``+``        ``(y ``-` `circle_y) ``*` `(y ``-` `circle_y) <``=` `rad ``*` `rad):``        ``return` `True``;``    ``else``:``        ``return` `False``;` `# Driver Code``x ``=` `1``;``y ``=` `1``;``circle_x ``=` `0``;``circle_y ``=` `1``;``rad ``=` `2``;``if``(isInside(circle_x, circle_y, rad, x, y)):``    ``print``(``"Inside"``);``else``:``    ``print``(``"Outside"``);` `# This code is contributed``# by mits.`

C#

 `// C# program to check if a point lies``// inside a circle or not``using` `System;` `class` `GFG {` `    ``static` `bool` `isInside(``int` `circle_x, ``int` `circle_y,``                              ``int` `rad, ``int` `x, ``int` `y)``    ``{``        ``// Compare radius of circle with``        ``// distance of its center from``        ``// given point``        ``if` `((x - circle_x) * (x - circle_x) +``            ``(y - circle_y) * (y - circle_y)    <= rad * rad)``            ``return` `true``;``        ``else``            ``return` `false``;``    ``}` `    ``// Driver Program to test above function``    ``public` `static` `void` `Main()``    ``{``        ``int` `x = 1, y = 1;``        ``int` `circle_x = 0, circle_y = 1, rad = 2;` `        ``if` `(isInside(circle_x, circle_y, rad, x, y))``            ``Console.Write(``"Inside"``);``        ``else``            ``Console.Write(``"Outside"``);``    ``}``}` `// This code is contributed by nitin mittal.`

PHP

 ``

Javascript

 ``

Output :

`Inside`

Time complexity: O(1) because doing constant operations

Auxiliary space: O(1)

Thanks to Utkarsh Trivedi for suggesting above solution