三点求圆心的算法是什么？

Question

我在圆周上有三个点：

 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;
  }

原文由 Russell Strauss 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

这可能是一个相当深入的计算。这里有一个简单的步骤：http: //paulbourke.net/geometry/circlesphere/ 。一旦你有了圆的方程，你可以简单地把它放在一个包含 H 和 K 的形式中。点 (h,k) 将是圆心。

（在链接处向下滚动一点以获取方程式）

原文由 Azmisov 发布，翻译遵循 CC BY-SA 3.0 许可协议

Answer 2

这是我的 Java 端口，当行列式以非常优雅的方式消失时躲避错误情况 IllegalArgumentException ，我处理“两点相距很远”或“点位于一条线上”条件的方法。此外，这会计算您的相交斜率方法不会执行的半径（并处理异常情况）。

 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));
}

原文由 andersoj 发布，翻译遵循 CC BY-SA 4.0 许可协议