使用克拉默法则进行三点定圆（二维）

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1.二维圆

  已知不共线的三个点，设其坐标为$(x_1,y_1)$、 $(x_2,y_2)$、$(x_3,y_3)$，圆的一般方程为：
$$A(x^2+y^2)+Bx+Cy+D=0\text{\hspace{1em}(1)}$$
将式(1)变形可得圆的标准方程为：
$$(x+\frac{B}{2A}{)}^2+(y+\frac{C}{2A}{)}^2=\frac{B^2+C^2-4AD}{4A^2}\text{\hspace{1em}(2)}$$
将三个已知点代入式(1)，可得关于$A,B,C,D$的齐次线性方程组：

$$\left[\begin{array}{cccc}{x}^2+y^2& x& y& 1\\ x_1^2+y_1^2& x_1& y_1& 1\\ x_2^2+y_2^2& x_2& y_2& 1\\ x_3^2+y_3^2& x_3& y_3& 1\end{array}\right]\cdot \left[\begin{array}{c}A\\ B\\ C\\ D\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right]\text{\hspace{1em}(3)}$$
在三点不共线的前提下，该齐次线性方程组有非零解，其等价于系数矩阵不满秩，即有：

$$\left|\begin{array}{cccc}{x}^2+y^2& x& y& 1\\ x_1^2+y_1^2& x_1& y_1& 1\\ x_2^2+y_2^2& x_2& y_2& 1\\ x_3^2+y_3^2& x_3& y_3& 1\end{array}\right|=0\text{\hspace{1em}(4)}$$
将式(4)展开，并与式(1)对比可得四个系数：
$$A=+\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|\text{\hspace{1em}(5)}$$

$$B=-\left|\begin{array}{ccc}{x}_1^2+y_1^2& y_1& 1\\ x_2^2+y_2^2& y_2& 1\\ x_3^2+y_3^2& y_3& 1\end{array}\right|\text{\hspace{1em}(6)}$$

$$C=+\left|\begin{array}{ccc}{x}_1^2+y_1^2& x_1& 1\\ x_2^2+y_2^2& x_2& 1\\ x_3^2+y_3^2& x_3& 1\end{array}\right|\text{\hspace{1em}(7)}$$

$$D=-\left|\begin{array}{ccc}{x}_1^2+y_1^2& x_1& y_1\\ x_2^2+y_2^2& x_2& y_2\\ x_3^2+y_3^2& x_3& y_3\end{array}\right|\text{\hspace{1em}(8)}$$
由式(2)可得圆心坐标$(x_c,y_c)$和半径$r$，即
$$\{\begin{array}{l}{x}_c=-\frac{B}{2A}\\ y_c=-\frac{C}{2A}\\ r=\sqrt{\frac{B^2+C^2-4AD}{4A^2}}\end{array}\text{\hspace{1em}(9)}$$

2.python代码

import numpy as np


def three_points_fit_circle(points):
    P1 = points[0]
    P2 = points[1]
    P3 = points[2]
    # 共线检查
    temp01 = P1 - P2
    temp02 = P3 - P2
    temp03 = np.cross(temp01, temp02)
    temp = (temp03 @ temp03) / (temp01 @ temp01) / (temp02 @ temp02)
    if temp < 10 ** -6:
        print('\t三点共线, 无法确定圆')
        return None
    A = np.ones((3, 3))
    A[0, 0] = P1[0]
    A[0, 1] = P1[1]
    A[1, 0] = P2[0]
    A[1, 1] = P2[1]
    A[2, 0] = P3[0]
    A[2, 1] = P3[1]

    B = np.ones((3, 3))
    B[0, 0] = P1[0] ** 2 + P1[1] ** 2
    B[0, 1] = P1[1]
    B[1, 0] = P2[0] ** 2 + P2[1] ** 2
    B[1, 1] = P2[1]
    B[2, 0] = P2[0] ** 2 + P2[1] ** 2
    B[2, 1] = P3[1]

    C = np.ones((3, 3))
    C[0, 0] = P1[0] ** 2 + P1[1] ** 2
    C[0, 1] = P1[0]
    C[1, 0] = P2[0] ** 2 + P2[1] ** 2
    C[1, 1] = P2[0]
    C[2, 0] = P2[0] ** 2 + P2[1] ** 2
    C[2, 1] = P3[0]

    D = np.ones((3, 3))
    D[0, 0] = P1[0] ** 2 + P1[1] ** 2
    D[0, 1] = P1[0]
    D[0, 2] = P1[1]
    D[1, 0] = P2[0] ** 2 + P2[1] ** 2
    D[1, 1] = P2[0]
    D[1, 2] = P2[1]
    D[2, 0] = P2[0] ** 2 + P2[1] ** 2
    D[2, 1] = P3[0]
    D[2, 2] = P3[1]

    A = +np.linalg.det(A)
    B = -np.linalg.det(B)
    C = +np.linalg.det(C)
    D = -np.linalg.det(D)

    Xc = -B / (2 * A)
    Yc = -C / (2 * A)

    r = np.sqrt((B * B + C * C - 4 * A * D) / (4 * A * A))

    return Xc, Yc, r

3.计算结果

圆心x坐标:14.558850728542366
圆心y坐标:11.80858513779518
圆半径:22.163390629231692