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。

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。

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];
    }
}

我根据chmike给出的答案为iOS创建了这个函数

+ (NSArray *)intersectionPointsOfCircleWithCenter:(CGPoint)center withRadius:(float)radius toLinePoint1:(CGPoint)p1 andLinePoint2:(CGPoint)p2
{
    NSMutableArray *intersectionPoints = [NSMutableArray array];

    float Ax = p1.x;
    float Ay = p1.y;
    float Bx = p2.x;
    float By = p2.y;
    float Cx = center.x;
    float Cy = center.y;
    float R = radius;


    // compute the euclidean distance between A and B
    float LAB = sqrt( pow(Bx-Ax, 2)+pow(By-Ay, 2) );

    // compute the direction vector D from A to B
    float Dx = (Bx-Ax)/LAB;
    float Dy = (By-Ay)/LAB;

    // Now the line equation is x = Dx*t + Ax, y = Dy*t + Ay with 0 <= t <= 1.

    // compute the value t of the closest point to the circle center (Cx, Cy)
    float t = Dx*(Cx-Ax) + Dy*(Cy-Ay);

    // This is the projection of C on the line from A to B.

    // compute the coordinates of the point E on line and closest to C
    float Ex = t*Dx+Ax;
    float Ey = t*Dy+Ay;

    // compute the euclidean distance from E to C
    float LEC = sqrt( pow(Ex-Cx, 2)+ pow(Ey-Cy, 2) );

    // test if the line intersects the circle
    if( LEC < R )
    {
        // compute distance from t to circle intersection point
        float dt = sqrt( pow(R, 2) - pow(LEC,2) );

        // compute first intersection point
        float Fx = (t-dt)*Dx + Ax;
        float Fy = (t-dt)*Dy + Ay;

        // compute second intersection point
        float Gx = (t+dt)*Dx + Ax;
        float Gy = (t+dt)*Dy + Ay;

        [intersectionPoints addObject:[NSValue valueWithCGPoint:CGPointMake(Fx, Fy)]];
        [intersectionPoints addObject:[NSValue valueWithCGPoint:CGPointMake(Gx, Gy)]];
    }

    // else test if the line is tangent to circle
    else if( LEC == R ) {
        // tangent point to circle is E
        [intersectionPoints addObject:[NSValue valueWithCGPoint:CGPointMake(Ex, Ey)]];
    }
    else {
        // line doesn't touch circle
    }

    return intersectionPoints;
}

如果你找到了圆心(因为它是3D的，我想你是指球体而不是圆)和直线之间的距离，然后检查这个距离是否小于可以做到这一点的半径。

碰撞点显然是直线和球面之间最近的点(当你计算球面和直线之间的距离时，会计算出这个点)

点与线之间的距离: http://mathworld.wolfram.com/Point-LineDistance3-Dimensional.html