我发现这个解决方案似乎比其他一些解决方案更容易遵循。

采取:

p1 and p2 as the points for the line, and
c as the center point for the circle and r for the radius

我可以用斜截式来解直线方程。但是，我不想处理以c为点的复杂方程，所以我只是平移了坐标系使圆在(0,0)处

p3 = p1 - c
p4 = p2 - c

顺便说一下，当我相互减分的时候，我是在减去x再减去y，然后把它们放到一个新的点里，以防有人不知道。

不管怎样，我现在解出p3和p4的直线方程

m = (p4_y - p3_y) / (p4_x - p3) (the underscore is an attempt at subscript)
y = mx + b
y - mx = b (just put in a point for x and y, and insert the m we found)

好的。现在我需要让这两个方程相等。首先我需要解圆的x方程

x^2 + y^2 = r^2
y^2 = r^2 - x^2
y = sqrt(r^2 - x^2)

然后我让它们相等:

mx + b = sqrt(r^2 - x^2)

求二次方程(0 = ax^2 + bx + c)

(mx + b)^2 = r^2 - x^2
(mx)^2 + 2mbx + b^2 = r^2 - x^2
0 = m^2 * x^2 + x^2 + 2mbx + b^2 - r^2
0 = (m^2 + 1) * x^2 + 2mbx + b^2 - r^2

现在我有了a b c。

a = m^2 + 1
b = 2mb
c = b^2 - r^2

我把这个代入二次公式

(-b ± sqrt(b^2 - 4ac)) / 2a

用值代入，然后尽可能简化:

(-2mb ± sqrt(b^2 - 4ac)) / 2a
(-2mb ± sqrt((-2mb)^2 - 4(m^2 + 1)(b^2 - r^2))) / 2(m^2 + 1)
(-2mb ± sqrt(4m^2 * b^2 - 4(m^2 * b^2 - m^2 * r^2 + b^2 - r^2))) / 2m^2 + 2
(-2mb ± sqrt(4 * (m^2 * b^2 - (m^2 * b^2 - m^2 * r^2 + b^2 - r^2))))/ 2m^2 + 2
(-2mb ± sqrt(4 * (m^2 * b^2 - m^2 * b^2 + m^2 * r^2 - b^2 + r^2)))/ 2m^2 + 2
(-2mb ± sqrt(4 * (m^2 * r^2 - b^2 + r^2)))/ 2m^2 + 2
(-2mb ± sqrt(4) * sqrt(m^2 * r^2 - b^2 + r^2))/ 2m^2 + 2
(-2mb ± 2 * sqrt(m^2 * r^2 - b^2 + r^2))/ 2m^2 + 2
(-2mb ± 2 * sqrt(m^2 * r^2 + r^2 - b^2))/ 2m^2 + 2
(-2mb ± 2 * sqrt(r^2 * (m^2 + 1) - b^2))/ 2m^2 + 2

这几乎是化简的极限了。最后，分离出带有±的方程:

(-2mb + 2 * sqrt(r^2 * (m^2 + 1) - b^2))/ 2m^2 + 2 or     
(-2mb - 2 * sqrt(r^2 * (m^2 + 1) - b^2))/ 2m^2 + 2 

然后简单地将这两个方程的结果代入mx + b中的x。为了清晰起见，我写了一些JavaScript代码来演示如何使用这个:

function interceptOnCircle(p1,p2,c,r){
    //p1 is the first line point
    //p2 is the second line point
    //c is the circle's center
    //r is the circle's radius

    var p3 = {x:p1.x - c.x, y:p1.y - c.y} //shifted line points
    var p4 = {x:p2.x - c.x, y:p2.y - c.y}

    var m = (p4.y - p3.y) / (p4.x - p3.x); //slope of the line
    var b = p3.y - m * p3.x; //y-intercept of line

    var underRadical = Math.pow((Math.pow(r,2)*(Math.pow(m,2)+1)),2)-Math.pow(b,2)); //the value under the square root sign 

    if (underRadical < 0){
    //line completely missed
        return false;
    } else {
        var t1 = (-2*m*b+2*Math.sqrt(underRadical))/(2 * Math.pow(m,2) + 2); //one of the intercept x's
        var t2 = (-2*m*b-2*Math.sqrt(underRadical))/(2 * Math.pow(m,2) + 2); //other intercept's x
        var i1 = {x:t1,y:m*t1+b} //intercept point 1
        var i2 = {x:t2,y:m*t2+b} //intercept point 2
        return [i1,i2];
    }
}

我希望这能有所帮助!

附注:如果任何人发现任何错误或有任何建议，请评论。我是新手，欢迎大家的帮助/建议。

另一种解决方案，首先考虑不关心碰撞位置的情况。请注意，这个特定的函数是在假设xB和yB为向量输入的情况下构建的，但如果情况并非如此，则可以轻松修改。变量名在函数的开头定义

#Line segment points (A0, Af) defined by xA0, yA0, xAf, yAf; circle center denoted by xB, yB; rB=radius of circle, rA = radius of point (set to zero for your application)
def staticCollision_f(xA0, yA0, xAf, yAf, rA, xB, yB, rB): #note potential speed up here by casting all variables to same type and/or using Cython
    
    #Build equations of a line for linear agents (convert y = mx + b to ax + by + c = 0 means that a = -m, b = 1, c = -b
    m_v = (yAf - yA0) / (xAf - xA0)
    b_v = yAf - m_v * xAf
    rEff = rA + rB #radii are added since we are considering the agent path as a thin line

    #Check if points (circles) are within line segment (find center of line segment and check if circle is within radius of this point)
    segmentMask = np.sqrt( (yB - (yA0+yAf)/2)**2 + (xB - (xA0+xAf)/2)**2 ) < np.sqrt( (yAf - yA0)**2 + (xAf - xA0)**2 ) / 2 + rEff

    #Calculate perpendicular distance between line and a point
    dist_v = np.abs(-m_v * xB + yB - b_v) / np.sqrt(m_v**2 + 1)
    collisionMask = (dist_v < rEff) & segmentMask

    #return True if collision is detected
    return collisionMask, collisionMask.any()

如果您需要碰撞的位置，您可以使用这个站点上详细介绍的方法，并将其中一个代理的速度设置为零。这种方法也适用于矢量输入:http://twobitcoder.blogspot.com/2010/04/circle-collision-detection.html

You can find a point on a infinite line that is nearest to circle center by projecting vector AC onto vector AB. Calculate the distance between that point and circle center. If it is greater that R, there is no intersection. If the distance is equal to R, line is a tangent of the circle and the point nearest to circle center is actually the intersection point. If distance less that R, then there are 2 intersection points. They lie at the same distance from the point nearest to circle center. That distance can easily be calculated using Pythagorean theorem. Here's algorithm in pseudocode:

{
dX = bX - aX;
dY = bY - aY;
if ((dX == 0) && (dY == 0))
  {
  // A and B are the same points, no way to calculate intersection
  return;
  }

dl = (dX * dX + dY * dY);
t = ((cX - aX) * dX + (cY - aY) * dY) / dl;

// point on a line nearest to circle center
nearestX = aX + t * dX;
nearestY = aY + t * dY;

dist = point_dist(nearestX, nearestY, cX, cY);

if (dist == R)
  {
  // line segment touches circle; one intersection point
  iX = nearestX;
  iY = nearestY;

  if (t < 0 || t > 1)
    {
    // intersection point is not actually within line segment
    }
  }
else if (dist < R)
  {
  // two possible intersection points

  dt = sqrt(R * R - dist * dist) / sqrt(dl);

  // intersection point nearest to A
  t1 = t - dt;
  i1X = aX + t1 * dX;
  i1Y = aY + t1 * dY;
  if (t1 < 0 || t1 > 1)
    {
    // intersection point is not actually within line segment
    }

  // intersection point farthest from A
  t2 = t + dt;
  i2X = aX + t2 * dX;
  i2Y = aY + t2 * dY;
  if (t2 < 0 || t2 > 1)
    {
    // intersection point is not actually within line segment
    }
  }
else
  {
  // no intersection
  }
}

编辑:增加了代码来检查所找到的交点是否实际上在线段内。

好吧，我不会给你代码，但既然你已经标记了这个算法，我认为这对你来说无关紧要。 首先，你要得到一个垂直于这条直线的向量。

y = ax + c是一个未知变量c是未知变量 为了解决这个问题，计算直线经过圆心时的值。

也就是说, 将圆心的位置代入直线方程，解出c。 然后计算原直线与其法线的交点。

这样就能得到直线上离圆最近的点。 计算该点到圆中心之间的距离(使用矢量的大小)。 如果这个小于圆的半径，看，我们有一个交点!

也许有另一种方法来解决这个问题，使用坐标系的旋转。

通常，如果一个线段是水平的或垂直的，这意味着平行于x轴或y轴，交点的求解很容易，因为我们已经知道交点的一个坐标，如果有的话。剩下的显然是用圆的方程找到另一个坐标。

受此启发，我们可以利用坐标系旋转，使一个轴的方向与线段的方向重合。

让我们以圆x^2+y^2=1和线段P1-P2为例，P1(-1.5,0.5)和P2(-0.5，-0.5)在x-y系统中。下面的方程提醒你旋转的原理，其中是逆时针方向的角度，x'-y'是旋转后的方程组:

x'=x*cos () + y*sin () y' = - x*sin () + y*cos ()

和反向

X = X ' * cos - y' * sin Y = x' * sin + Y ' * cos

考虑P1-P2方向(用-x表示为45°)，我们可以取=45°。将第二个旋转方程转化为x-y系统中的圆方程:x^2+y^2=1，经过简单的运算，我们得到x'-y'系统中的“相同”方程:x'^2+y'^2=1。

利用第一个旋转方程=> P1(-根号(2)/2，根号(2))，P2(-根号(2)/ 2,0)，线段端点变成x'-y'系统。

假设交点为p，在x'-y'中，Px = -根号2 /2。使用新的圆方程，我们得到Py = +根号(2)/2。将P转换成原始的x-y系统，最终得到P(-1,0)

为了实现这个数值，我们可以先看看线段的方向:水平，垂直或不垂直。如果它属于前两种情况，很简单。如果是最后一种情况，应用上述算法。

为了判断是否有交集，我们可以将解与端点坐标进行比较，看看它们之间是否有一个根。

我相信只要我们有了它的方程，这个方法也可以应用于其他曲线。唯一的缺点是，我们应该在x'-y'坐标系下解方程，这可能很难。

另一个在c#(部分圆类)。 经过测试，工作就像一个魅力。

public class Circle : IEquatable<Circle>
{
    // ******************************************************************
    // The center of a circle
    private Point _center;
    // The radius of a circle
    private double _radius;

   // ******************************************************************
    /// <summary>
    /// Find all intersections (0, 1, 2) of the circle with a line defined by its 2 points.
    /// Using: http://math.stackexchange.com/questions/228841/how-do-i-calculate-the-intersections-of-a-straight-line-and-a-circle
    /// Note: p is the Center.X and q is Center.Y
    /// </summary>
    /// <param name="linePoint1"></param>
    /// <param name="linePoint2"></param>
    /// <returns></returns>
    public List<Point> GetIntersections(Point linePoint1, Point linePoint2)
    {
        List<Point> intersections = new List<Point>();

        double dx = linePoint2.X - linePoint1.X;

        if (dx.AboutEquals(0)) // Straight vertical line
        {
            if (linePoint1.X.AboutEquals(Center.X - Radius) || linePoint1.X.AboutEquals(Center.X + Radius))
            {
                Point pt = new Point(linePoint1.X, Center.Y);
                intersections.Add(pt);
            }
            else if (linePoint1.X > Center.X - Radius && linePoint1.X < Center.X + Radius)
            {
                double x = linePoint1.X - Center.X;

                Point pt = new Point(linePoint1.X, Center.Y + Math.Sqrt(Radius * Radius - (x * x)));
                intersections.Add(pt);

                pt = new Point(linePoint1.X, Center.Y - Math.Sqrt(Radius * Radius - (x * x)));
                intersections.Add(pt);
            }

            return intersections;
        }

        // Line function (y = mx + b)
        double dy = linePoint2.Y - linePoint1.Y;
        double m = dy / dx;
        double b = linePoint1.Y - m * linePoint1.X;

        double A = m * m + 1;
        double B = 2 * (m * b - m * _center.Y - Center.X);
        double C = Center.X * Center.X + Center.Y * Center.Y - Radius * Radius - 2 * b * Center.Y + b * b;

        double discriminant = B * B - 4 * A * C;

        if (discriminant < 0)
        {
            return intersections; // there is no intersections
        }

        if (discriminant.AboutEquals(0)) // Tangeante (touch on 1 point only)
        {
            double x = -B / (2 * A);
            double y = m * x + b;

            intersections.Add(new Point(x, y));
        }
        else // Secant (touch on 2 points)
        {
            double x = (-B + Math.Sqrt(discriminant)) / (2 * A);
            double y = m * x + b;
            intersections.Add(new Point(x, y));

            x = (-B - Math.Sqrt(discriminant)) / (2 * A);
            y = m * x + b;
            intersections.Add(new Point(x, y));
        }

        return intersections;
    }

    // ******************************************************************
    // Get the center
    [XmlElement("Center")]
    public Point Center
    {
        get { return _center; }
        set
        {
            _center = value;
        }
    }

    // ******************************************************************
    // Get the radius
    [XmlElement]
    public double Radius
    {
        get { return _radius; }
        set { _radius = value; }
    }

    //// ******************************************************************
    //[XmlArrayItemAttribute("DoublePoint")]
    //public List<Point> Coordinates
    //{
    //    get { return _coordinates; }
    //}

    // ******************************************************************
    // Construct a circle without any specification
    public Circle()
    {
        _center.X = 0;
        _center.Y = 0;
        _radius = 0;
    }

    // ******************************************************************
    // Construct a circle without any specification
    public Circle(double radius)
    {
        _center.X = 0;
        _center.Y = 0;
        _radius = radius;
    }

    // ******************************************************************
    // Construct a circle with the specified circle
    public Circle(Circle circle)
    {
        _center = circle._center;
        _radius = circle._radius;
    }

    // ******************************************************************
    // Construct a circle with the specified center and radius
    public Circle(Point center, double radius)
    {
        _center = center;
        _radius = radius;
    }

    // ******************************************************************
    // Construct a circle based on one point
    public Circle(Point center)
    {
        _center = center;
        _radius = 0;
    }

    // ******************************************************************
    // Construct a circle based on two points
    public Circle(Point p1, Point p2)
    {
        Circle2Points(p1, p2);
    }

要求:

using System;

namespace Mathematic
{
    public static class DoubleExtension
    {
        // ******************************************************************
        // Base on Hans Passant Answer on:
        // http://stackoverflow.com/questions/2411392/double-epsilon-for-equality-greater-than-less-than-less-than-or-equal-to-gre

        /// <summary>
        /// Compare two double taking in account the double precision potential error.
        /// Take care: truncation errors accumulate on calculation. More you do, more you should increase the epsilon.
        public static bool AboutEquals(this double value1, double value2)
        {
            if (double.IsPositiveInfinity(value1))
                return double.IsPositiveInfinity(value2);

            if (double.IsNegativeInfinity(value1))
                return double.IsNegativeInfinity(value2);

            if (double.IsNaN(value1))
                return double.IsNaN(value2);

            double epsilon = Math.Max(Math.Abs(value1), Math.Abs(value2)) * 1E-15;
            return Math.Abs(value1 - value2) <= epsilon;
        }

        // ******************************************************************
        // Base on Hans Passant Answer on:
        // http://stackoverflow.com/questions/2411392/double-epsilon-for-equality-greater-than-less-than-less-than-or-equal-to-gre

        /// <summary>
        /// Compare two double taking in account the double precision potential error.
        /// Take care: truncation errors accumulate on calculation. More you do, more you should increase the epsilon.
        /// You get really better performance when you can determine the contextual epsilon first.
        /// </summary>
        /// <param name="value1"></param>
        /// <param name="value2"></param>
        /// <param name="precalculatedContextualEpsilon"></param>
        /// <returns></returns>
        public static bool AboutEquals(this double value1, double value2, double precalculatedContextualEpsilon)
        {
            if (double.IsPositiveInfinity(value1))
                return double.IsPositiveInfinity(value2);

            if (double.IsNegativeInfinity(value1))
                return double.IsNegativeInfinity(value2);

            if (double.IsNaN(value1))
                return double.IsNaN(value2);

            return Math.Abs(value1 - value2) <= precalculatedContextualEpsilon;
        }

        // ******************************************************************
        public static double GetContextualEpsilon(this double biggestPossibleContextualValue)
        {
            return biggestPossibleContextualValue * 1E-15;
        }

        // ******************************************************************
        /// <summary>
        /// Mathlab equivalent
        /// </summary>
        /// <param name="dividend"></param>
        /// <param name="divisor"></param>
        /// <returns></returns>
        public static double Mod(this double dividend, double divisor)
        {
            return dividend - System.Math.Floor(dividend / divisor) * divisor;
        }

        // ******************************************************************
    }
}