PHP

if ((($x - $center_x) ** 2 + ($y - $center_y) ** 2) <=  $radius **2) {
    return true; // Inside
} else {
    return false; // Outside
}

我在c#中的回答是一个完整的剪切和粘贴(未优化)解决方案:

public static bool PointIsWithinCircle(double circleRadius, double circleCenterPointX, double circleCenterPointY, double pointToCheckX, double pointToCheckY)
{
    return (Math.Pow(pointToCheckX - circleCenterPointX, 2) + Math.Pow(pointToCheckY - circleCenterPointY, 2)) < (Math.Pow(circleRadius, 2));
}

用法:

if (!PointIsWithinCircle(3, 3, 3, .5, .5)) { }

下面是解决这个问题的简单java代码:

以及它背后的数学:https://math.stackexchange.com/questions/198764/how-to-know-if-a-point-is-inside-a-circle

boolean insideCircle(int[] point, int[] center, int radius) {
    return (float)Math.sqrt((int)Math.pow(point[0]-center[0],2)+(int)Math.pow(point[1]-center[1],2)) <= radius;
}

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;
}

这样效率更高，可读性更强。它避免了昂贵的平方根运算。我还添加了一个检查，以确定点是否在圆的边界矩形内。

矩形检查是不必要的，除非有许多点或许多圆。如果大多数点都在圆圈内，边框检查实际上会使事情变慢!

像往常一样，一定要考虑您的用例。

进入3D世界，如果你想检查一个3D点是否在单位球面上，你最终会做类似的事情。在2D中工作所需要的就是使用2D矢量运算。

    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;
    }

下面的方程是一个表达式，测试一个点是否在一个给定的圆内，其中xP和yP是点的坐标，xC和yC是圆心的坐标，R是给定圆的半径。

如果上述表达式为真，则该点在圆内。

下面是一个c#实现的示例:

    public static bool IsWithinCircle(PointF pC, Point pP, Single fRadius){
        return Distance(pC, pP) <= fRadius;
    }

    public static Single Distance(PointF p1, PointF p2){
        Single dX = p1.X - p2.X;
        Single dY = p1.Y - p2.Y;
        Single multi = dX * dX + dY * dY;
        Single dist = (Single)Math.Round((Single)Math.Sqrt(multi), 3);

        return (Single)dist;
    }