我知道自从这个帖子被打开以来已经有一段时间了。根据chmike给出的答案，经Aqib Mumtaz改进。他们给出了一个很好的答案，但只适用于无限线，就像Aqib说的那样。所以我添加了一些比较来知道线段是否与圆接触，我用Python写的。

def LineIntersectCircle(c, r, p1, p2):
    #p1 is the first line point
    #p2 is the second line point
    #c is the circle's center
    #r is the circle's radius

    p3 = [p1[0]-c[0], p1[1]-c[1]]
    p4 = [p2[0]-c[0], p2[1]-c[1]]

    m = (p4[1] - p3[1]) / (p4[0] - p3[0])
    b = p3[1] - m * p3[0]

    underRadical = math.pow(r,2)*math.pow(m,2) + math.pow(r,2) - math.pow(b,2)

    if (underRadical < 0):
        print("NOT")
    else:
        t1 = (-2*m*b+2*math.sqrt(underRadical)) / (2 * math.pow(m,2) + 2)
        t2 = (-2*m*b-2*math.sqrt(underRadical)) / (2 * math.pow(m,2) + 2)
        i1 = [t1+c[0], m * t1 + b + c[1]]
        i2 = [t2+c[0], m * t2 + b + c[1]]

        if p1[0] > p2[0]:                                           #Si el punto 1 es mayor al 2 en X
            if (i1[0] < p1[0]) and (i1[0] > p2[0]):                 #Si el punto iX esta entre 2 y 1 en X
                if p1[1] > p2[1]:                                   #Si el punto 1 es mayor al 2 en Y
                    if (i1[1] < p1[1]) and (i1[1] > p2[1]):         #Si el punto iy esta entre 2 y 1
                        print("Intersection")
                if p1[1] < p2[1]:                                   #Si el punto 2 es mayo al 2 en Y
                    if (i1[1] > p1[1]) and (i1[1] < p2[1]):         #Si el punto iy esta entre 1 y 2
                        print("Intersection")

        if p1[0] < p2[0]:                                           #Si el punto 2 es mayor al 1 en X
            if (i1[0] > p1[0]) and (i1[0] < p2[0]):                 #Si el punto iX esta entre 1 y 2 en X
                if p1[1] > p2[1]:                                   #Si el punto 1 es mayor al 2 en Y
                    if (i1[1] < p1[1]) and (i1[1] > p2[1]):         #Si el punto iy esta entre 2 y 1
                        print("Intersection")
                if p1[1] < p2[1]:                                   #Si el punto 2 es mayo al 2 en Y
                    if (i1[1] > p1[1]) and (i1[1] < p2[1]):         #Si el punto iy esta entre 1 y 2
                        print("Intersection")

        if p1[0] > p2[0]:                                           #Si el punto 1 es mayor al 2 en X
            if (i2[0] < p1[0]) and (i2[0] > p2[0]):                 #Si el punto iX esta entre 2 y 1 en X
                if p1[1] > p2[1]:                                   #Si el punto 1 es mayor al 2 en Y
                    if (i2[1] < p1[1]) and (i2[1] > p2[1]):         #Si el punto iy esta entre 2 y 1
                        print("Intersection")
                if p1[1] < p2[1]:                                   #Si el punto 2 es mayo al 2 en Y
                    if (i2[1] > p1[1]) and (i2[1] < p2[1]):         #Si el punto iy esta entre 1 y 2
                        print("Intersection")

        if p1[0] < p2[0]:                                           #Si el punto 2 es mayor al 1 en X
            if (i2[0] > p1[0]) and (i2[0] < p2[0]):                 #Si el punto iX esta entre 1 y 2 en X
                if p1[1] > p2[1]:                                   #Si el punto 1 es mayor al 2 en Y
                    if (i2[1] < p1[1]) and (i2[1] > p2[1]):         #Si el punto iy esta entre 2 y 1
                        print("Intersection")
                if p1[1] < p2[1]:                                   #Si el punto 2 es mayo al 2 en Y
                    if (i2[1] > p1[1]) and (i2[1] < p2[1]):         #Si el punto iy esta entre 1 y 2
                        print("Intersection")

这里你需要一些数学知识:

假设A = (Xa, Ya)， B = (Xb, Yb)， C = (Xc, Yc)。从A到B的直线上的任意一点都有坐标(*Xa + (1-)Xb， * ya + (1-)*Yb) = P

如果点P的距离是R到C，它一定在圆上。你想要的是解决

distance(P, C) = R

这是

(alpha*Xa + (1-alpha)*Xb)^2 + (alpha*Ya + (1-alpha)*Yb)^2 = R^2
alpha^2*Xa^2 + alpha^2*Xb^2 - 2*alpha*Xb^2 + Xb^2 + alpha^2*Ya^2 + alpha^2*Yb^2 - 2*alpha*Yb^2 + Yb^2=R^2
(Xa^2 + Xb^2 + Ya^2 + Yb^2)*alpha^2 - 2*(Xb^2 + Yb^2)*alpha + (Xb^2 + Yb^2 - R^2) = 0

如果你将abc公式应用到这个方程来求解，并使用alpha的解来计算P的坐标，你会得到交点，如果存在的话。

另一种方法使用三角形ABC面积公式。交点检验比投影法简单高效，但求交点坐标需要更多的工作。至少它会被推迟到需要的时候。

三角形面积的计算公式为:area = bh/2

b是底长，h是高。我们选择线段AB作为底，使h是圆心C到直线的最短距离。

因为三角形的面积也可以用向量点积来计算，所以我们可以确定h。

// compute the triangle area times 2 (area = area2/2)
area2 = abs( (Bx-Ax)*(Cy-Ay) - (Cx-Ax)(By-Ay) )

// compute the AB segment length
LAB = sqrt( (Bx-Ax)² + (By-Ay)² )

// compute the triangle height
h = area2/LAB

// if the line intersects the circle
if( h < R )
{
    ...
}        

更新1:

您可以通过使用这里描述的快速平方根倒数计算来优化代码，以获得1/LAB的良好近似值。

计算交点并不难。开始了

// compute the line AB direction vector components
Dx = (Bx-Ax)/LAB
Dy = (By-Ay)/LAB

// compute the distance from A toward B of closest point to C
t = Dx*(Cx-Ax) + Dy*(Cy-Ay)

// t should be equal to sqrt( (Cx-Ax)² + (Cy-Ay)² - h² )

// compute the intersection point distance from t
dt = sqrt( R² - h² )

// compute first intersection point coordinate
Ex = Ax + (t-dt)*Dx
Ey = Ay + (t-dt)*Dy

// compute second intersection point coordinate
Fx = Ax + (t+dt)*Dx
Fy = Ay + (t+dt)*Dy

如果h = R，则直线AB与圆相切，且值dt = 0, E = F。点的坐标为E和F的坐标。

如果在应用程序中出现这种情况，您应该检查A与B是否不同，并且段长度不为空。

如果直线的坐标为A.x, A.y和B.x, B.y，圆心为C.x, C.y，则直线公式为:

x = A.x * t + B.x * (1 - t)

y = A.y * t + B.y * (1 - t)

0 < = t < = 1

这个圆是

(C.x - x)²+ (C.y - y)²= R²

如果你把直线的x和y公式代入圆公式，你会得到一个t的二阶方程，它的解是交点(如果有的话)。如果你得到的t小于0或大于1，那么它不是一个解，但它表明这条线“指向”圆的方向。

奇怪的是，我可以回答，但不能评论…… 我喜欢Multitaskpro的方法，它可以移动所有东西，使圆的中心落在原点上。不幸的是，他的代码中有两个问题。首先在平方根下的部分，你需要去掉双倍的幂。所以不是:

is underRadical = Math.pow（（Math.pow（r，2）*（Math.pow（m，2）+1）），2）-Math.pow（b，2））;

but:

under Radical = Math.pow（r，2）*（Math.pow（m，2）+1）） - Math.pow（b，2）;

在最后的坐标中，他忘记把解移回来。所以不是:

var i1 = {x：t1，y：m*t1+b}

but:

Var i1 = {x:t1+c。x, y: m * t1 + b +陈守惠};

整个函数就变成:

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

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

这是一个Javascript实现。我的方法是首先将线段转换成一条无限的直线，然后找到交点。从那里，我检查是否找到的点在线段上。代码有良好的文档记录，您应该能够跟随。

您可以在这个现场演示中试用代码。 代码是从我的算法仓库里拿的。

// Small epsilon value
var EPS = 0.0000001;

// point (x, y)
function Point(x, y) {
  this.x = x;
  this.y = y;
}

// Circle with center at (x,y) and radius r
function Circle(x, y, r) {
  this.x = x;
  this.y = y;
  this.r = r;
}

// A line segment (x1, y1), (x2, y2)
function LineSegment(x1, y1, x2, y2) {
  var d = Math.sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) );
  if (d < EPS) throw 'A point is not a line segment';
  this.x1 = x1; this.y1 = y1;
  this.x2 = x2; this.y2 = y2;
}

// An infinite line defined as: ax + by = c
function Line(a, b, c) {
  this.a = a; this.b = b; this.c = c;
  // Normalize line for good measure
  if (Math.abs(b) < EPS) {
    c /= a; a = 1; b = 0;
  } else { 
    a = (Math.abs(a) < EPS) ? 0 : a / b;
    c /= b; b = 1; 
  }
}

// Given a line in standard form: ax + by = c and a circle with 
// a center at (x,y) with radius r this method finds the intersection
// of the line and the circle (if any). 
function circleLineIntersection(circle, line) {

  var a = line.a, b = line.b, c = line.c;
  var x = circle.x, y = circle.y, r = circle.r;

  // Solve for the variable x with the formulas: ax + by = c (equation of line)
  // and (x-X)^2 + (y-Y)^2 = r^2 (equation of circle where X,Y are known) and expand to obtain quadratic:
  // (a^2 + b^2)x^2 + (2abY - 2ac + - 2b^2X)x + (b^2X^2 + b^2Y^2 - 2bcY + c^2 - b^2r^2) = 0
  // Then use quadratic formula X = (-b +- sqrt(a^2 - 4ac))/2a to find the 
  // roots of the equation (if they exist) and this will tell us the intersection points

  // In general a quadratic is written as: Ax^2 + Bx + C = 0
  // (a^2 + b^2)x^2 + (2abY - 2ac + - 2b^2X)x + (b^2X^2 + b^2Y^2 - 2bcY + c^2 - b^2r^2) = 0
  var A = a*a + b*b;
  var B = 2*a*b*y - 2*a*c - 2*b*b*x;
  var C = b*b*x*x + b*b*y*y - 2*b*c*y + c*c - b*b*r*r;

  // Use quadratic formula x = (-b +- sqrt(a^2 - 4ac))/2a to find the 
  // roots of the equation (if they exist).

  var D = B*B - 4*A*C;
  var x1,y1,x2,y2;

  // Handle vertical line case with b = 0
  if (Math.abs(b) < EPS) {

    // Line equation is ax + by = c, but b = 0, so x = c/a
    x1 = c/a;

    // No intersection
    if (Math.abs(x-x1) > r) return [];

    // Vertical line is tangent to circle
    if (Math.abs((x1-r)-x) < EPS || Math.abs((x1+r)-x) < EPS)
      return [new Point(x1, y)];

    var dx = Math.abs(x1 - x);
    var dy = Math.sqrt(r*r-dx*dx);

    // Vertical line cuts through circle
    return [
      new Point(x1,y+dy),
      new Point(x1,y-dy)
    ];

  // Line is tangent to circle
  } else if (Math.abs(D) < EPS) {

    x1 = -B/(2*A);
    y1 = (c - a*x1)/b;

    return [new Point(x1,y1)];

  // No intersection
  } else if (D < 0) {

    return [];

  } else {

    D = Math.sqrt(D);

    x1 = (-B+D)/(2*A);
    y1 = (c - a*x1)/b;

    x2 = (-B-D)/(2*A);
    y2 = (c - a*x2)/b;

    return [
      new Point(x1, y1),
      new Point(x2, y2)
    ];

  }

}

// Converts a line segment to a line in general form
function segmentToGeneralForm(x1,y1,x2,y2) {
  var a = y1 - y2;
  var b = x2 - x1;
  var c = x2*y1 - x1*y2;
  return new Line(a,b,c);
}

// Checks if a point 'pt' is inside the rect defined by (x1,y1), (x2,y2)
function pointInRectangle(pt,x1,y1,x2,y2) {
  var x = Math.min(x1,x2), X = Math.max(x1,x2);
  var y = Math.min(y1,y2), Y = Math.max(y1,y2);
  return x - EPS <= pt.x && pt.x <= X + EPS &&
         y - EPS <= pt.y && pt.y <= Y + EPS;
}

// Finds the intersection(s) of a line segment and a circle
function lineSegmentCircleIntersection(segment, circle) {

  var x1 = segment.x1, y1 = segment.y1, x2 = segment.x2, y2 = segment.y2;
  var line = segmentToGeneralForm(x1,y1,x2,y2);
  var pts = circleLineIntersection(circle, line);

  // No intersection
  if (pts.length === 0) return [];

  var pt1 = pts[0];
  var includePt1 = pointInRectangle(pt1,x1,y1,x2,y2);

  // Check for unique intersection
  if (pts.length === 1) {
    if (includePt1) return [pt1];
    return [];
  }

  var pt2 = pts[1];
  var includePt2 = pointInRectangle(pt2,x1,y1,x2,y2);

  // Check for remaining intersections
  if (includePt1 && includePt2) return [pt1, pt2];
  if (includePt1) return [pt1];
  if (includePt2) return [pt2];
  return [];

}