从三点求圆心的算法是什么？

Question

我在圆的圆周上有三个点：

pt A = (A.x, A.y);
pt B = (B.x, B.y);
pt C = (C.x, C.y);

如何计算圆心？

在Processing (Java) 中实现它。

我找到了答案并实施了一个可行的解决方案：

 pt circleCenter(pt A, pt B, pt C) {

    float yDelta_a = B.y - A.y;
    float xDelta_a = B.x - A.x;
    float yDelta_b = C.y - B.y;
    float xDelta_b = C.x - B.x;
    pt center = P(0,0);

    float aSlope = yDelta_a/xDelta_a;
    float bSlope = yDelta_b/xDelta_b;  
    center.x = (aSlope*bSlope*(A.y - C.y) + bSlope*(A.x + B.x)
        - aSlope*(B.x+C.x) )/(2* (bSlope-aSlope) );
    center.y = -1*(center.x - (A.x+B.x)/2)/aSlope +  (A.y+B.y)/2;

    return center;
  }

Answer 1

这是我的 Java 端口，当行列式消失时，用一个非常优雅的 IllegalArgumentException 来避免错误条件，这是我处理“两个点相距很远”或“点位于一条线上”条件的方法。此外，这会计算半径（并应对特殊条件），而您的相交斜坡方法将无法做到这一点。

public class CircleThree
{ 
  static final double TOL = 0.0000001;
  
  public static Circle circleFromPoints(final Point p1, final Point p2, final Point p3)
  {
    final double offset = Math.pow(p2.x,2) + Math.pow(p2.y,2);
    final double bc =   ( Math.pow(p1.x,2) + Math.pow(p1.y,2) - offset )/2.0;
    final double cd =   (offset - Math.pow(p3.x, 2) - Math.pow(p3.y, 2))/2.0;
    final double det =  (p1.x - p2.x) * (p2.y - p3.y) - (p2.x - p3.x)* (p1.y - p2.y); 
    
    if (Math.abs(det) < TOL) { throw new IllegalArgumentException("Yeah, lazy."); }

    final double idet = 1/det;
     
    final double centerx =  (bc * (p2.y - p3.y) - cd * (p1.y - p2.y)) * idet;
    final double centery =  (cd * (p1.x - p2.x) - bc * (p2.x - p3.x)) * idet;
    final double radius = 
       Math.sqrt( Math.pow(p2.x - centerx,2) + Math.pow(p2.y-centery,2));
    
    return new Circle(new Point(centerx,centery),radius);
  }
  
  static class Circle
  {
    final Point center;
    final double radius;
    public Circle(Point center, double radius)
    {
      this.center = center; this.radius = radius;
    }
    @Override 
    public String toString()
    {
      return new StringBuilder().append("Center= ").append(center).append(", r=").append(radius).toString();
    }
  }
  
  static class Point
  {
    final double x,y;

    public Point(double x, double y)
    {
      this.x = x; this.y = y;
    }
    @Override
    public String toString()
    {
      return "("+x+","+y+")";
    }
    
  }

  public static void main(String[] args)
  {
    Point p1 = new Point(0.0,1.0);
    Point p2 = new Point(1.0,0.0);
    Point p3 = new Point(2.0,1.0);
    Circle c = circleFromPoints(p1, p2, p3);
    System.out.println(c);
  }
  
}

请参阅数学论坛的算法：

void circle_vvv(circle *c)
{
    c->center.w = 1.0;
    vertex *v1 = (vertex *)c->c.p1;
    vertex *v2 = (vertex *)c->c.p2;
    vertex *v3 = (vertex *)c->c.p3;
    float bx = v1->xw; float by = v1->yw;
    float cx = v2->xw; float cy = v2->yw;
    float dx = v3->xw; float dy = v3->yw;
    float temp = cx*cx+cy*cy;
    float bc = (bx*bx + by*by - temp)/2.0;
    float cd = (temp - dx*dx - dy*dy)/2.0;
    float det = (bx-cx)*(cy-dy)-(cx-dx)*(by-cy);
    if (fabs(det) < 1.0e-6) {
        c->center.xw = c->center.yw = 1.0;
        c->center.w = 0.0;
        c->v1 = *v1;
        c->v2 = *v2;
        c->v3 = *v3;
        return;
        }
    det = 1/det;
    c->center.xw = (bc*(cy-dy)-cd*(by-cy))*det;
    c->center.yw = ((bx-cx)*cd-(cx-dx)*bc)*det;
    cx = c->center.xw; cy = c->center.yw;
    c->radius = sqrt((cx-bx)*(cx-bx)+(cy-by)*(cy-by));
}

Answer 2

这可能是一个相当深入的计算。这里有一个简单的分步步骤： http://paulbourke.net/geometry/circlesphere/。一旦你有了圆的方程，你可以简单地把它写成包含 H 和 K 的形式。点 (h,k) 将是圆心。

（在链接处向下滚动一点即可找到方程式）

Answer 3

当我将鼠标悬停在这个问题上时，我正在寻找类似的算法。
获取了您的代码，但发现这在斜率为 0 或无穷大的情况下不起作用（当 xDelta_a 或 xDelta_b 为 0 时可能为真）。

我更正了算法，这是我的代码。
注意：我使用objective-c编程语言，只是更改点值初始化的代码，所以如果这是错误的，我相信使用java工作的程序员可以纠正它。然而，所有的逻辑都是相同的（上帝保佑算法!!:)）

就我自己的功能测试而言，工作得很好。
如果逻辑有任何错误，请告诉我。

pt circleCenter(pt A, pt B, pt C) {

float yDelta_a = B.y - A.y;
float xDelta_a = B.x - A.x;
float yDelta_b = C.y - B.y;
float xDelta_b = C.x - B.x;
pt center = P(0,0);

float aSlope = yDelta_a/xDelta_a;
float bSlope = yDelta_b/xDelta_b;

pt AB_Mid = P((A.x+B.x)/2, (A.y+B.y)/2);
pt BC_Mid = P((B.x+C.x)/2, (B.y+C.y)/2);

if(yDelta_a == 0)         //aSlope == 0
{
    center.x = AB_Mid.x;
    if (xDelta_b == 0)         //bSlope == INFINITY
    {
        center.y = BC_Mid.y;
    }
    else
    {
        center.y = BC_Mid.y + (BC_Mid.x-center.x)/bSlope;
    }
}
else if (yDelta_b == 0)               //bSlope == 0
{
    center.x = BC_Mid.x;
    if (xDelta_a == 0)             //aSlope == INFINITY
    {
        center.y = AB_Mid.y;
    }
    else
    {
        center.y = AB_Mid.y + (AB_Mid.x-center.x)/aSlope;
    }
}
else if (xDelta_a == 0)        //aSlope == INFINITY
{
    center.y = AB_Mid.y;
    center.x = bSlope*(BC_Mid.y-center.y) + BC_Mid.x;
}
else if (xDelta_b == 0)        //bSlope == INFINITY
{
    center.y = BC_Mid.y;
    center.x = aSlope*(AB_Mid.y-center.y) + AB_Mid.x;
}
else
{
    center.x = (aSlope*bSlope*(AB_Mid.y-BC_Mid.y) - aSlope*BC_Mid.x + bSlope*AB_Mid.x)/(bSlope-aSlope);
    center.y = AB_Mid.y - (center.x - AB_Mid.x)/aSlope;
}

return center;
}

Answer 4

很抱歉我的回答迟了。当两个点形成一条垂直线时，任何使用“斜率”的解决方案都会失败，因为斜率将是无限的。

这是 2019 年的一个简单稳健的解决方案，始终正常工作：

public static boolean circleCenter(double[] p1, double[] p2, double[] p3, double[] center) {
    double ax = (p1[0] + p2[0]) / 2;
    double ay = (p1[1] + p2[1]) / 2;
    double ux = (p1[1] - p2[1]);
    double uy = (p2[0] - p1[0]);
    double bx = (p2[0] + p3[0]) / 2;
    double by = (p2[1] + p3[1]) / 2;
    double vx = (p2[1] - p3[1]);
    double vy = (p3[0] - p2[0]);
    double dx = ax - bx;
    double dy = ay - by;
    double vu = vx * uy - vy * ux;
    if (vu == 0)
        return false; // Points are collinear, so no unique solution
    double g = (dx * uy - dy * ux) / vu;
    center[0] = bx + g * vx;
    center[1] = by + g * vy;
    return true;
}

当且仅当 3 个点共线时，上述代码才会返回“false”。

Answer 5

public Vector2 CarculateCircleCenter(Vector2 p1, Vector2 p2, Vector2 p3)
{
    if (
        p2.x - p1.x == 0 ||
        p3.x - p2.x == 0 ||
        p2.y - p1.y == 0 ||
        p3.y - p2.y == 0
    ) return null;

    Vector2 center = new Vector2();
    float ma = (p2.y - p1.y) / (p2.x - p1.x);
    float mb = (p3.y - p2.y) / (p3.x - p2.x);
    center.x = (ma * mb * (p1.y - p3.y) + mb * (p1.x - p2.x) - ma * (p2.x + p3.x)) / (2 * (mb - ma));
    center.y = (-1 / ma) * (center.x - (p1.x + p2.x) * 0.5) + (p1.y + p2.y) * 0.5;
    return center;
}

Answer 6

def circle_mid_point(list_b):
list_k = []
for i in range(3):
    for j in range(1,4):
        if i+j <=3:
            midpoint_x1 =  (list_b[i][0] + list_b[i+j][0])/2
            midpoint_y1 = (list_b[i][1] + list_b[i + j][1]) / 2
            list_k.append([midpoint_x1,midpoint_y1]) # list of all the midpoints of the lines

for k in range(len(list_k)):
    for j in range(1,len(list_k)-1):
        if list_k[k] == list_k[k+j]: #at centre only two midpoints will have the same value
            return list_k[k]

k = circle_mid_point([[0,1],[1,0],[-1,0],[0,-1]])
print(k)

Answer 7

了解Circumcircle

特殊情况Apollonius 问题：类型 1：三点

求外心和半径：

local function ppp (x1, y1, x2, y2, x3, y3)
    local d = 2 * (x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2))
    local t1, t2, t3 = x1*x1+y1*y1, x2*x2+y2*y2, x3*x3+y3*y3
    local x = (t1 * (y2 - y3) + t2 * (y3 - y1) + t3 * (y1 - y2)) / d
    local y = (t1 * (x3 - x2) + t2 * (x1 - x3) + t3 * (x2 - x1)) / d
    local radius = math.sqrt((x1-x)^2 + (y1-y)^2)
    return x, y, radius
end

示例（结果：x、y、半径）：

print (ppp(0, 0, 0, 100, 300, 300)) -- 250, 50, 254.95097567964
print (ppp(0, 100, 250, 50, 300, 300)) -- 150, 200, 180.2775637732