Equation for testing if a point is inside a circle

Question

If you have a circle with center (center_x, center_y) and radius radius, how do you test if a given point with coordinates (x, y) is inside the circle?

Answer 1

boolean isInRectangle(double centerX, double centerY, double radius, 
    double x, double y)
{
        return x >= centerX - radius && x <= centerX + radius && 
            y >= centerY - radius && y <= centerY + radius;
}    

//test if coordinate (x, y) is within a radius from coordinate (center_x, center_y)
public boolean isPointInCircle(double centerX, double centerY, 
    double radius, double x, double y)
{
    if(isInRectangle(centerX, centerY, radius, x, y))
    {
        double dx = centerX - x;
        double dy = centerY - y;
        dx *= dx;
        dy *= dy;
        double distanceSquared = dx + dy;
        double radiusSquared = radius * radius;
        return distanceSquared <= radiusSquared;
    }
    return false;
}

This is more efficient, and readable. It avoids the costly square root operation. I also added a check to determine if the point is within the bounding rectangle of the circle.

The rectangle check is unnecessary except with many points or many circles. If most points are inside circles, the bounding rectangle check will actually make things slower!

As always, be sure to consider your use case.

Answer 2

Calculate the Distance

D = Math.Sqrt(Math.Pow(center_x - x, 2) + Math.Pow(center_y - y, 2))
return D <= radius

that's in C#...convert for use in python...

Answer 3

I used the code below for beginners like me :).

public class incirkel {

public static void main(String[] args) {
    int x; 
    int y; 
    int middelx; 
    int middely; 
    int straal; {

// Adjust the coordinates of x and y 
x = -1;
y = -2;

// Adjust the coordinates of the circle
middelx = 9; 
middely = 9;
straal =  10;

{
    //When x,y is within the circle the message below will be printed
    if ((((middelx - x) * (middelx - x)) 
                    + ((middely - y) * (middely - y))) 
                    < (straal * straal)) {
                        System.out.println("coordinaten x,y vallen binnen cirkel");
    //When x,y is NOT within the circle the error message below will be printed
    } else {
        System.err.println("x,y coordinaten vallen helaas buiten de cirkel");
    } 
}



    }
}}

Answer 4

This is the same solution as mentioned by Jason Punyon, but it contains a pseudo-code example and some more details. I saw his answer after writing this, but I didn't want to remove mine.

I think the most easily understandable way is to first calculate the distance between the circle's center and the point. I would use this formula:

d = sqrt((circle_x - x)^2 + (circle_y - y)^2)

Then, simply compare the result of that formula, the distance (d), with the radius. If the distance (d) is less than or equal to the radius (r), the point is inside the circle (on the edge of the circle if d and r are equal).

Here is a pseudo-code example which can easily be converted to any programming language:

function is_in_circle(circle_x, circle_y, r, x, y)
{
    d = sqrt((circle_x - x)^2 + (circle_y - y)^2);
    return d <= r;
}

Where circle_x and circle_y is the center coordinates of the circle, r is the radius of the circle, and x and y is the coordinates of the point.

Answer 5

You should check whether the distance from the center of the circle to the point is smaller than the radius, i.e.

if (x-center_x)**2 + (y-center_y)**2 <= radius**2:
    # inside circle

Answer 6

Moving into the world of 3D if you want to check if a 3D point is in a Unit Sphere you end up doing something similar. All that is needed to work in 2D is to use 2D vector operations.

    public static bool Intersects(Vector3 point, Vector3 center, float radius)
    {
        Vector3 displacementToCenter = point - center;

        float radiusSqr = radius * radius;

        bool intersects = displacementToCenter.magnitude < radiusSqr;

        return intersects;
    }