' VB.NET - Code

Function CheckLineSegmentCircleIntersection(x1 As Double, y1 As Double, x2 As Double, y2 As Double, xc As Double, yc As Double, r As Double) As Boolean
    Static xd As Double = 0.0F
    Static yd As Double = 0.0F
    Static t As Double = 0.0F
    Static d As Double = 0.0F
    Static dx_2_1 As Double = 0.0F
    Static dy_2_1 As Double = 0.0F

    dx_2_1 = x2 - x1
    dy_2_1 = y2 - y1

    t = ((yc - y1) * dy_2_1 + (xc - x1) * dx_2_1) / (dy_2_1 * dy_2_1 + dx_2_1 * dx_2_1)

    If 0 <= t And t <= 1 Then
        xd = x1 + t * dx_2_1
        yd = y1 + t * dy_2_1

        d = Math.Sqrt((xd - xc) * (xd - xc) + (yd - yc) * (yd - yc))
        Return d <= r
    Else
        d = Math.Sqrt((xc - x1) * (xc - x1) + (yc - y1) * (yc - y1))
        If d <= r Then
            Return True
        Else
            d = Math.Sqrt((xc - x2) * (xc - x2) + (yc - y2) * (yc - y2))
            If d <= r Then
                Return True
            Else
                Return False
            End If
        End If
    End If
End Function

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。

这个Java函数返回一个DVec2对象。它用DVec2表示圆心，用DVec2表示半径，用Line表示直线。

public static DVec2 CircLine(DVec2 C, double r, Line line)
{
    DVec2 A = line.p1;
    DVec2 B = line.p2;
    DVec2 P;
    DVec2 AC = new DVec2( C );
    AC.sub(A);
    DVec2 AB = new DVec2( B );
    AB.sub(A);
    double ab2 = AB.dot(AB);
    double acab = AC.dot(AB);
    double t = acab / ab2;

    if (t < 0.0) 
        t = 0.0;
    else if (t > 1.0) 
        t = 1.0;

    //P = A + t * AB;
    P = new DVec2( AB );
    P.mul( t );
    P.add( A );

    DVec2 H = new DVec2( P );
    H.sub( C );
    double h2 = H.dot(H);
    double r2 = r * r;

    if(h2 > r2) 
        return null;
    else
        return P;
}

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。

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

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!