已知三个点的坐标，怎样计算其圆心的坐标及半径？

[OneCoderr]

已知三个点的坐标，怎样计算其圆心的坐标及半径？ 不是用直尺画，而是用C#写出代码

[Hobo丶]

其实计算圆心和坐标 只不过把数学的用代码展现出来 一样的

[欢乐的小猪]

 class Program
    {
        static void Main(string[] args)
        {
            CircleInfo circle = new CircleInfo(new CPoint() { X = -2, Y = 0 }, new CPoint() { X = 0, Y = 2 }, new CPoint() { X = 2, Y = 0 });
            Console.WriteLine(string.Format("CenterX:{0}\tCenterY:{1}\tRadius:{2}", circle.Center.X, circle.Center.Y, circle.Radius));
            Console.ReadKey();
        }
    }
    class CircleInfo
    {
        public CircleInfo(CPoint p1, CPoint p2, CPoint p3)
        {
            this.center = new CPoint();
            if ((p2.X - p1.X) * (p3.Y - p1.Y) == (p2.Y - p1.Y) * (p3.X - p1.X))//三点在一条直线上
            {
                ;
            }
            else
            {
                if (p2.Y - p1.Y != 0 && p3.Y - p1.Y != 0)
                {
                    SetCircleInfo(p1, p2, p3);
                }
                else if (p1.Y - p2.Y != 0 && p3.Y - p2.Y != 0)
                {
                    SetCircleInfo(p2, p1, p3);
                }
                else if (p2.Y - p3.Y != 0 && p1.Y - p3.Y != 0)
                {
                    SetCircleInfo(p3, p2, p1);
                }
            }
        }


        private CPoint center;

        public CPoint Center
        {
            get { return center; }
            set { center = value; }
        }


        private double radius;

        public double Radius
        {
            get { return radius; }
            set { radius = value; }
        }

        void SetCircleInfo(CPoint p1, CPoint p2, CPoint p3)
        {
            double ratio1 = -(p2.X - p1.X) / (p2.Y - p1.Y);
            double ratio2 = -(p3.X - p1.X) / (p3.Y - p1.Y);
            CPoint cp1 = new CPoint() { X = 0.5 * (p1.X + p2.X), Y = 0.5 * (p1.Y + p2.Y) };
            CPoint cp2 = new CPoint() { X = 0.5 * (p1.X + p3.X), Y = 0.5 * (p1.Y + p3.Y) };
            center.X = (cp2.Y - cp1.Y + ratio1 * cp1.X - ratio2 * cp2.X) / (ratio1 - ratio2);
            center.Y = ratio1 * (center.X - cp1.X) + cp1.Y;

            radius = Math.Pow((p1.X - center.X) * (p1.X - center.X) + (p1.Y - center.Y) * (p1.Y - center.Y), 0.5);
        }
    }

    class CPoint
    {
        public double X { get; set; }
        public double Y { get; set; }
    }

[白衣如花]

中垂线的交点。 两点组成一条线段，线段可做中垂线，两中垂线有交点，即圆心 圆心到三点距离是半径

[白衣如花]

斜率 (ya - yb) / (xa - xb) 中点 ((xa + xb) / 2, (ya + yb) / 2) 垂线斜率 = -1/斜率 = -(xa - xb) / (ya - yb) 所以中垂线方程： y - (ya + yb) / 2 = ( -(xa - xb) / (ya - yb)) * (x - (xa + xb) / 2) 两直线交点，就是解二元一次方程，自己想吧

[巴士上的邂逅]

A(x1,y1), B(x2,y2),C(x3,y3), AB的中点M（ (x1+x2)/2, (y1+y2)/2) AB的斜率：k=(y2-y1)/(x2-x1) AB的垂直平分线的斜率为:-1/k 因此由点斜式写出AB垂直平分线：y=-(x2-x1/(y2-y1)* [x-(x1+x2)/2]+(y1+y2)/2 同理得到AC垂直平分线：y=-(x3-x1/(y3-y1)* [x-(x1+x3)/2]+(y1+y3)/2 两垂直平分线交点就是圆心，二元一次方程应该会吧，得出的x、y就是圆心坐标 A、B、C任意一点到圆心的距离就是半径 啊！！我是来教数学的！

[良朋]

            
            double u = (y[1] - y[0]) / (x[1] - x[0]);
            double v = (y[2] - y[1]) / (x[2] - x[1]);
            double w = 0.5 * (y[0] + y[1]) + 0.5 * (x[0] + x[1]) / ((y[1] - y[0]) / (x[1] - x[0]));
            double k = 0.5 * (y[1] + y[2]) + 0.5 * (x[1] + x[2]) / ((y[2] - y[1]) / (x[2] - x[1]));

            double cx = (k - w) / (1 / v - 1 / u);
            double cy = (w - k) / u / (1 / v - 1 / u) + w;
            double r = Math.Sqrt(Math.Pow((y[0] - cy), 2) + Math.Pow((x[0] - cx), 2));