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

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。

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; ｝

' 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

我写了一个小脚本，通过将圆的中心点投影到直线上来测试相交。

vector distVector = centerPoint - projectedPoint;
if(distVector.length() < circle.radius)
{
    double distance = circle.radius - distVector.length();
    vector moveVector = distVector.normalize() * distance;
    circle.move(moveVector);
}

http://jsfiddle.net/ercang/ornh3594/1/

如果需要检查与线段的碰撞，还需要考虑圆心到起点和终点的距离。

vector distVector = centerPoint - startPoint;
if(distVector.length() < circle.radius)
{
    double distance = circle.radius - distVector.length();
    vector moveVector = distVector.normalize() * distance;
    circle.move(moveVector);
}

https://jsfiddle.net/ercang/menp0991/

圆真的是一个坏人:)所以一个好办法是避免真正的圆，如果可以的话。如果你正在为游戏做碰撞检查，你可以进行一些简化，只做3个点积，并进行一些比较。

我称之为“胖点”或“瘦圈”。它是平行于线段方向上半径为0的椭圆。而是垂直于线段方向的全半径

首先，我会考虑重命名和切换坐标系统，以避免过多的数据:

s0s1 = B-A;
s0qp = C-A;
rSqr = r*r;

其次，hvec2f中的索引h意味着vector必须支持水平操作，如dot()/det()。这意味着它的组件被放置在一个单独的xmm寄存器中，以避免shuffle /hadd'ing/hsub'ing。现在我们开始，最简单的2D游戏碰撞检测的最佳性能版本:

bool fat_point_collides_segment(const hvec2f& s0qp, const hvec2f& s0s1, const float& rSqr) {
    auto a = dot(s0s1, s0s1);
    //if( a != 0 ) // if you haven't zero-length segments omit this, as it would save you 1 _mm_comineq_ss() instruction and 1 memory fetch
    {
        auto b = dot(s0s1, s0qp);
        auto t = b / a; // length of projection of s0qp onto s0s1
        //std::cout << "t = " << t << "\n";
        if ((t >= 0) && (t <= 1)) // 
        {
            auto c = dot(s0qp, s0qp);
            auto r2 = c - a * t * t;
            return (r2 <= rSqr); // true if collides
        }
    }   
    return false;
}

我怀疑你能进一步优化它。我正在用它进行神经网络驱动的赛车碰撞检测，处理数百万个迭代步骤。