下面是JavaScript的一个很好的解决方案(包括所有必需的数学和实时插图) https://bl.ocks.org/milkbread/11000965

尽管该解决方案中的is_on函数需要修改:

函数is_on(a, b, c) { return Math.abs(距离(a,c) +距离(c,b) -距离(a,b))<0.000001; ｝

只是这个线程的一个补充… 下面是pahlevan发布的代码版本，但针对c# /XNA，并做了一些整理:

    /// <summary>
    /// Intersects a line and a circle.
    /// </summary>
    /// <param name="location">the location of the circle</param>
    /// <param name="radius">the radius of the circle</param>
    /// <param name="lineFrom">the starting point of the line</param>
    /// <param name="lineTo">the ending point of the line</param>
    /// <returns>true if the line and circle intersect each other</returns>
    public static bool IntersectLineCircle(Vector2 location, float radius, Vector2 lineFrom, Vector2 lineTo)
    {
        float ab2, acab, h2;
        Vector2 ac = location - lineFrom;
        Vector2 ab = lineTo - lineFrom;
        Vector2.Dot(ref ab, ref ab, out ab2);
        Vector2.Dot(ref ac, ref ab, out acab);
        float t = acab / ab2;

        if (t < 0)
            t = 0;
        else if (t > 1)
            t = 1;

        Vector2 h = ((ab * t) + lineFrom) - location;
        Vector2.Dot(ref h, ref h, out h2);

        return (h2 <= (radius * radius));
    }

虽然我认为使用线圆交点，然后检查交点是否在端点之间更好，可能更便宜，但我想添加这个更直观的解决方案。

我喜欢把这个问题想象成“香肠上的点问题”，在不改变算法的情况下，它可以在任何维度上工作。 这个解找不到交点。

以下是我想到的:

(我使用“小于”，但“小于或等于”也可以使用，这取决于我们测试的内容。)

确保Circle_Point小于到无限线的半径距离。(这里使用最喜欢的方法)。 计算从两个Segment_Points到Circle_Point的距离。 测试较大的Circle_Point-Segment_Point距离是否小于根号(Segment_Length^2+Radius^2)。 (这是从一个分段点到一个理论点的距离，也就是从另一个分段点到无限线(直角)的半径距离。见图片)。

3 t。如果为true: Circle_Point在sausage内部。 3 f。如果为false:如果较小的Circle_Point- segment_point距离小于Radius，则Circle_Point在sausage内部。

图片:最粗的线段是选定的线段，没有示例圆。有点粗糙，有些像素有点不对。

function boolean pointInSausage(sp1,sp2,r,c) {
  if ( !(pointLineDist(c,sp1,sp2) < r) ) {
    return false;
  }
  double a = dist(sp1,c);  
  double b = dist(sp2,c);
  double l;
  double s;
  if (a>b) {
    l = a;
    s = b;
  } else {
    l = b;
    s = a;
  }
  double segLength = dist(sp1,sp2);
  if ( l < sqrt(segLength*segLength+r*r) ) {
    return true;
  }
  return s < r;
}  

如果发现任何问题，告诉我，我会编辑或撤回。

以下是我在TypeScript中的解决方案，遵循@Mizipzor建议的想法(使用投影):

/**
 * Determines whether a line segment defined by a start and end point intersects with a sphere defined by a center point and a radius
 * @param a the start point of the line segment
 * @param b the end point of the line segment
 * @param c the center point of the sphere
 * @param r the radius of the sphere
 */
export function lineSphereIntersects(
  a: IPoint,
  b: IPoint,
  c: IPoint,
  r: number
): boolean {
  // find the three sides of the triangle formed by the three points
  const ab: number = distance(a, b);
  const ac: number = distance(a, c);
  const bc: number = distance(b, c);

  // check to see if either ends of the line segment are inside of the sphere
  if (ac < r || bc < r) {
    return true;
  }

  // find the angle between the line segment and the center of the sphere
  const numerator: number = Math.pow(ac, 2) + Math.pow(ab, 2) - Math.pow(bc, 2);
  const denominator: number = 2 * ac * ab;
  const cab: number = Math.acos(numerator / denominator);

  // find the distance from the center of the sphere and the line segment
  const cd: number = Math.sin(cab) * ac;

  // if the radius is at least as long as the distance between the center and the line
  if (r >= cd) {
    // find the distance between the line start and the point on the line closest to
    // the center of the sphere
    const ad: number = Math.cos(cab) * ac;
    // intersection occurs when the point on the line closest to the sphere center is
    // no further away than the end of the line
    return ad <= ab;
  }
  return false;
}

export function distance(a: IPoint, b: IPoint): number {
  return Math.sqrt(
    Math.pow(b.z - a.z, 2) + Math.pow(b.y - a.y, 2) + Math.pow(b.x - a.x, 2)
  );
}

export interface IPoint {
  x: number;
  y: number;
  z: number;
}

采取

E是射线的起点， L是射线的端点， C是你测试的圆心 R是球面的半径

计算: d = L - E(射线方向矢量，从头到尾) f = E - C(从中心球到射线起点的向量)

然后通过…找到交点。 堵塞: P = E + t * d 这是一个参数方程 Px = Ex + tdx Py = Ey + tdy 成 (x - h)2 + (y - k)2 = r2 (h,k) =圆心。

注意:我们在这里将问题简化为2D，我们得到的解决方案也适用于3D

得到:

Expand x2 - 2xh + h2 + y2 - 2yk + k2 - r2 = 0 Plug x = ex + tdx y = ey + tdy ( ex + tdx )2 - 2( ex + tdx )h + h2 + ( ey + tdy )2 - 2( ey + tdy )k + k2 - r2 = 0 Explode ex2 + 2extdx + t2dx2 - 2exh - 2tdxh + h2 + ey2 + 2eytdy + t2dy2 - 2eyk - 2tdyk + k2 - r2 = 0 Group t2( dx2 + dy2 ) + 2t( exdx + eydy - dxh - dyk ) + ex2 + ey2 - 2exh - 2eyk + h2 + k2 - r2 = 0 Finally, t2( d · d ) + 2t( e · d - d · c ) + e · e - 2( e · c ) + c · c - r2 = 0 Where d is the vector d and · is the dot product. And then, t2( d · d ) + 2t( d · ( e - c ) ) + ( e - c ) · ( e - c ) - r2 = 0 Letting f = e - c t2( d · d ) + 2t( d · f ) + f · f - r2 = 0

所以我们得到: T2 *(d·d) + 2t*(f·d) + (f·f - r2) = 0

求解二次方程:

float a = d.Dot( d ) ;
float b = 2*f.Dot( d ) ;
float c = f.Dot( f ) - r*r ;

float discriminant = b*b-4*a*c;
if( discriminant < 0 )
{
  // no intersection
}
else
{
  // ray didn't totally miss sphere,
  // so there is a solution to
  // the equation.
  
  discriminant = sqrt( discriminant );

  // either solution may be on or off the ray so need to test both
  // t1 is always the smaller value, because BOTH discriminant and
  // a are nonnegative.
  float t1 = (-b - discriminant)/(2*a);
  float t2 = (-b + discriminant)/(2*a);

  // 3x HIT cases:
  //          -o->             --|-->  |            |  --|->
  // Impale(t1 hit,t2 hit), Poke(t1 hit,t2>1), ExitWound(t1<0, t2 hit), 

  // 3x MISS cases:
  //       ->  o                     o ->              | -> |
  // FallShort (t1>1,t2>1), Past (t1<0,t2<0), CompletelyInside(t1<0, t2>1)
  
  if( t1 >= 0 && t1 <= 1 )
  {
    // t1 is the intersection, and it's closer than t2
    // (since t1 uses -b - discriminant)
    // Impale, Poke
    return true ;
  }

  // here t1 didn't intersect so we are either started
  // inside the sphere or completely past it
  if( t2 >= 0 && t2 <= 1 )
  {
    // ExitWound
    return true ;
  }
  
  // no intn: FallShort, Past, CompletelyInside
  return false ;
}

我会用这个算法来计算点(圆心)和线(线AB)之间的距离。这可以用来确定直线与圆的交点。

假设有点A B c, Ax和Ay是A点的x和y分量。B和c也是一样，标量R是圆半径。

该算法要求A B C是不同的点，且R不为0。

这是算法

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

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

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

// compute the distance between the points A and E, where
// E is the point of AB closest the circle center (Cx, Cy)
t = Dx*(Cx-Ax) + Dy*(Cy-Ay)    

// compute the coordinates of the point E
Ex = t*Dx+Ax
Ey = t*Dy+Ay

// compute the euclidean distance between E and C
LEC = sqrt((Ex-Cx)²+(Ey-Cy)²)

// test if the line intersects the circle
if( LEC < R )
{
    // compute distance from t to circle intersection point
    dt = sqrt( R² - LEC²)

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

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

// else test if the line is tangent to circle
else if( LEC == R )
    // tangent point to circle is E

else
    // line doesn't touch circle