Find if a point lies inside a Circle

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

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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++

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// C++ program to check if a point 
// lies inside a circle or not
#include <bits/stdc++.h>
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";
}

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Java

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// 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.

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?php
// PHP program to check if a point 
// lies inside a circle or not
  
function 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)) 
echo "Inside";
else
echo "Outside";
  
// This code is contributed
// by nitin mittal.
?>

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Output :

Inside

Thanks to Utkarsh Trivedi for suggesting above solution

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



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Improved By : nitin mittal, Mithun Kumar



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