已知圆心上的三点： $\tiny (a_1,b_1)$ ， $\tiny (a_2,b_2)$ ， $\tiny (a_3,b_3)$ ，求圆心： $\tiny O(R_x,R_y)$

把三点坐标圆的方程： $\tiny (x-Rx)^2+(y-Ry)^2=r^2$ 得：

$\tiny (a_1-Rx)^2+(b_1-Ry)^2=(a_2-Rx)^2+(b_2-Ry)^2 \dots \dots \dots (1)$

$\tiny (a_1-Rx)^2+(b_1-Ry)^2=(a_3-Rx)^2+(b_3-Ry)^2 \dots \dots \dots (2)$

把 $\tiny (1)$ 展开整理得：

$\tiny a_1^2-a_2^2+b_1^2-b_2^2=(2a_1-2a_2)R_x+(2b_1-2b_2)R_y\dots \dots \dots (3)$

$\tiny a_1^2-a_3^2+b_1^2-b_3^2=(2a_1-2a_3)R_x+(2b_1-2b_3)R_y\dots \dots \dots (4)$

令

$\tiny A = a_1^2-a_2^2+b_1^2-b_2^2$ ;
$\tiny B = a_1^2-a_3^2+b_1^2-b_3^2$ ;
$\tiny P_1=b_1-b_3$ , $\tiny P_2=b_1-b_2$ ,
$\tiny Q_1=a_1-a_3$ , $\tiny Q_2=a_1-a_2$

$\tiny A=2Q_2R_x+2P_2R_y\dots \dots \dots (5)$

$\tiny B=2Q_1R_x+2P_1R_y\dots \dots \dots (6)$

求 $\tiny R_y$

$\tiny Q_1A=2Q_1Q_2R_x+2Q_1P_2R_y\dots \dots \dots (7)$

$\tiny Q_2B=2Q_1Q_2R_x+2P_1Q_2R_y\dots \dots \dots (8)$

求 $\tiny R_x$

$\tiny P_1A=2P_1Q_2R_x+2P_1P_2R_y\dots \dots \dots (9)$

$\tiny P_2B=2P_2Q_1R_x+2P_1P_2R_y\dots \dots \dots (10)$

把 $\tiny (7)-(8)$ 移项整理得：

$\tiny Q_1A-Q_2B=(2P_2Q_1-2P_1Q_2)R_y\dots \dots \dots (11)$

$\tiny R_y = \frac{Q_1A-Q_2B}{2P_2Q_1-2P_1Q_2}\dots \dots \dots (12)$

把 $\tiny (9)-(10)$ 移项整理得：

$\tiny P_1A-P_2B=(2P_1Q_2-2P_2Q_1)R_x\dots \dots \dots (13)$

$\tiny R_x = \frac{P_1A-P_2B}{2Q_2P_1-2Q_1P_2}\dots \dots \dots (14)$

还原 $\tiny (12)(14)$ 代入式子：

$\tiny R_x=\frac{(b_1-b_3)(b_1^2-b_2^2+a_1^2-a_2^2)-(b_1-b_2)(b_1^2-b_3^2+a_1^2-a_3^2)}{2(b_1-b_3)(a_1-a_2)-2(b_1-b_2)(a_1-a_3)}$

$\tiny R_y=\frac{(a_1-a_3)(a_1^2-a_2^2+b_1^2-b_2^2)-(a_1-a_2)(a_1^2-a_3^2+b_1^2-b_3^2)}{2(a_1-a_3)(b_1-b_2)-2(a_1-a_2)(b_1-b_3)}$

整理结果

$\tiny R_x=\frac{(b_1-b_3)(a_1^2-a_2^2+b_1^2-b_2^2)-(b_1-b_2)(a_1^2-a_3^2+b_1^2-b_3^2)}{2(a_1-a_2)(b_1-b_3)-2(a_1-a_3)(b_1-b_2)}$

$\tiny R_y=-\frac{(a_1-a_3)(a_1^2-a_2^2+b_1^2-b_2^2)-(a_1-a_2)(a_1^2-a_3^2+b_1^2-b_3^2)}{2(a_1-a_2)(b_1-b_3)-2(a_1-a_3)(b_1-b_2)}$

简化：

$\tiny R_x=\frac{P_1A-P_2B}{C}$

$\tiny R_y=-\frac{Q_1A-Q_2B}{C}$

其中：

$\tiny O(R_x,R_y)$ 圆心坐标
$\tiny P_1(a_1,b_1)$ ， $\tiny P_2(a_2,b_2)$ ， $\tiny P_3(a_3,b_3)$ 为不在同一直线上的坐标；
$\tiny A = a_1^2-a_2^2+b_1^2-b_2^2$ ;
$\tiny B = a_1^2-a_3^2+b_1^2-b_3^2$ ;
$\tiny C = 2(a_1-a_2)(b_1-b_3)-2(a_1-a_3)(b_1-b_2)$
$\tiny P_1=b_1-b_3$ , $\tiny P_2=b_1-b_2$ ,
$\tiny Q_1=a_1-a_3$ , $\tiny Q_2=a_1-a_2$

C#中的运用

        /// <summary>
        /// <para>二维：已知圆上三点，求圆心坐标</para>
        /// <para>三个点不在同一直线上</para>
        /// </summary>
        /// <param name="P1">点1</param>
        /// <param name="P2">点2</param>
        /// <param name="P3">点3</param>
        /// <returns></returns>
        public PointF GetCentre(PointF P1,PointF P2, PointF P3)
        {
            double a13 = P1.X - P3.X;
            double a13_ = P1.X + P3.X;
            double b13 = P1.Y - P3.Y;
            double b13_ = P1.Y + P3.Y;
            double a12 = P1.X - P2.X;
            double a12_ = P1.X + P2.X;
            double b12 = P1.Y - P2.Y;
            double b12_ = P1.Y + P2.Y;

            double a12b12_2 = a12 * a12_ + b12 * b12_;
            double a13b13_2 = a13 * a13_ + b13 * b13_;

            double a13b12 =2* a13 * b12;
            double a12b13 =2* a12 * b13;


            if (a12b13 - a13b12 == 0) return new PointF((P2.X+P1.X)/2, (P2.Y+P1.Y)/2);
            double af = a12b13 - a13b12;
            double bf = a13b12 - a12b13;
            double az = b13 * a12b12_2 - b12 * a13b13_2;
            double bz = a13 * a12b12_2 - a12 * a13b13_2;
            double a = az / af;
            double b = bz / bf;
            return new PointF((float)a,(float)b);
        }

少莫千华

发布了384 篇原创文章 · 获赞 70 · 访问量 68万+

他的留言板关注

【C#】已知圆心上的三点求圆心

整理结果

C#中的运用

猜你喜欢