Using Cramer's Rule for Four-Point Fixtures

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1. Algorithm process

  Four points that are not coplanar in the known space, let their coordinates be $A(x_1,y_1,z_1)$、 $B(x_2,y_2,z_2)$、$C(x_3,y_3,z_3)、D(x_4,y_4,z_4)$ , let the radius be$r$ , sphere center$O$ is$(x,y,z)$ . Using the property that the distances from the four points to the center of the sphere are equal, the following four equations are obtained.
$$\begin{gathered} (x-x_1{)}^2+(y-y_1{)}^2+(z-z_1{)}^2=r^2; \\ (x-x_2{)}^2+(y-y_2{)}^2+(z-z_2{)}^2=r^2; \\ (x-x_3{)}^2+(y-y_3{)}^2+(z-z_3{)}^2=r^2; \\ (x-x_4{)}^2+(y-y_4{)}^2+(z-z_4{)}^2=r^2; \end{gathered}$$

展开得:
$$\begin{gathered} x^2+y^2+z^2-2(x_{1}x+y_{1}y+z_{1}z)+x_1^2+y_1^2+z_1^2=r^2① \\ {x}^2+y^2+z^2-2(x_{2}x+y_{2}y+z_{2}z)+x_2^2+y_2^2+z_2^2=r^2② \\ {x}^2+y^2+z^2-2(x_{3}x+y_{3}y+z_{3}z)+x_3^2+y_3^2+z_3^2=r^2③ \\ {x}^2+y^2+z^2-2(x_{4}x+y_{4}y+z_{4}z)+x_4^2+y_4^2+z_4^2=r^2④ \end{gathered}$$

分别作①-②、③ - ④、② - ③得：
$$\begin{gathered} (x_1-x_2)x+(y_1-y_2)y+(z_1-z_2)z=1/2(x_1^2-x_2^2+y_1^2-y_2^2+z_1^2-z_2^2) \\ (x_3-x_4)x+(y_3-y_4)y+(z_3-z_4)z=1/2(x_3^2-x_4^2+y_3^2-y_4^2+z_3^2-z_4^2) \\ (x_2-x_3)x+(y_2-y_3)y+(z_2-z_3)z=1/2(x_2^2-x_3^2+y_2^2-y_3^2+z_2^2-z_3^2) \end{gathered}$$

Its corresponding coefficient determinant can be set as:

$$D=\left|\begin{array}{ccc}a& b& c\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$$

则：$$\begin{gathered} a=(x_1-x_2),b=(y_1-y_2),c=(z_1-z_2), \\ {a}_1=(x_3-x_4),b_1=(y_3-y_4)，c_1=(z_3-z_4), \\ {a}_2=(x_2-x_3),b_2=(y_2-y_3)，c_2=(z_2-z_3) \end{gathered}$$

The constant term determinant is:

$$L=\left|\begin{array}{c}P\\ Q\\ R\end{array}\right|$$

则：
$$P=\frac{1}{2}(x_1^2-x_2^2+y_1^2-y_2^2+z_1^2-z_2^2)$$
$$Q=\frac{1}{2}(x_3^2-x_4^2+y_3^2-y_4^2+z_3^2-z_4^2)$$
$$R=\frac{1}{2}(x_2^2-x_3^2+y_2^2-y_3^2+z_2^2-z_3^2)$$

现设：
$$Dx=\left|\begin{array}{ccc}P& b& c\\ Q& b_1& c_1\\ R& b_2& c_2\end{array}\right|$$

$$Dy=\left|\begin{array}{ccc}a& P& c\\ a_1& Q& c_1\\ a_2& R& c_2\end{array}\right|$$

$$Dz=\left|\begin{array}{ccc}a& b& P\\ a_1& b_1& Q\\ a_2& b_2& R\end{array}\right|$$

From Cramer's rule in linear algebra:
$$x=\frac{Dx}{D}$$

$$y=\frac{Dy}{D}$$

$$z=\frac{Dz}{D}$$

2. C++ code

bool calculatesphere_oAndsphere_r(Eigen::Vector3d& p1, Eigen::Vector3d& p2,
	Eigen::Vector3d& p3, Eigen::Vector3d& p4,
	Eigen::Vector3d& sphere_o, double& sphere_r)
{
    
    
	double a = p1[0] - p2[0], b = p1[1] - p2[1], c = p1[2] - p2[2];
	double a1 = p3[0] - p4[0], b1 = p3[1] - p4[1], c1 = p3[2] - p4[2];
	double a2 = p2[0] - p3[0], b2 = p2[1] - p3[1], c2 = p2[2] - p3[2];
	double A = p1[0] * p1[0] - p2[0] * p2[0];
	double B = p1[1] * p1[1] - p2[1] * p2[1];
	double C = p1[2] * p1[2] - p2[2] * p2[2];
	double A1 = p3[0] * p3[0] - p4[0] * p4[0];
	double B1 = p3[1] * p3[1] - p4[1] * p4[1];
	double C1 = p3[2] * p3[2] - p4[2] * p4[2];
	double A2 = p2[0] * p2[0] - p3[0] * p3[0];
	double B2 = p2[1] * p2[1] - p3[1] * p3[1];
	double C2 = p2[2] * p2[2] - p3[2] * p3[2];
	double P = (A + B + C) / 2;
	double Q = (A1 + B1 + C1) / 2;
	double R = (A2 + B2 + C2) / 2;

	// D是系数行列式，利用克拉默法则
	double D = a * b1 * c2 + a2 * b * c1 + c * a1 * b2 - (a2 * b1 * c + a1 * b * c2 + a * b2 * c1);

	double Dx = P * b1 * c2 + b * c1 * R + c * Q * b2 - (c * b1 * R + P * c1 * b2 + Q * b * c2);
	double Dy = a * Q * c2 + P * c1 * a2 + c * a1 * R - (c * Q * a2 + a * c1 * R + c2 * P * a1);
	double Dz = a * b1 * R + b * Q * a2 + P * a1 * b2 - (a2 * b1 * P + a * Q * b2 + R * b * a1);

	if (D == 0)
	{
    
    
		cerr << "四点共面" << endl;
		return false;
	}
	else
	{
    
    
		sphere_o[0] = Dx / D;
		sphere_o[1] = Dy / D;
		sphere_o[2] = Dz / D;
		sphere_r = sqrt((p1[0] - sphere_o[0]) * (p1[0] - sphere_o[0]) +
			(p1[1] - sphere_o[1]) * (p1[1] - sphere_o[1]) +
			(p1[2] - sphere_o[2]) * (p1[2] - sphere_o[2]));
		return true;
	}
}

3.python code

def four_point_fix_sphere(points):
    # 四个点的坐标
    P1 = points[0]
    P2 = points[1]
    P3 = points[2]
    P4 = points[3]
    # 互相相减
    l = P1 - P2
    l1 = P3 - P4
    l2 = P2 - P3
    # 系数D
    D = np.ones((3, 3))
    D[0, :] = l
    D[1, :] = l1
    D[2, :] = l2
    detD = np.linalg.det(D)
    # P,Q,R
    P = (np.linalg.norm(P1) ** 2 - np.linalg.norm(P2) ** 2) * 0.5
    Q = (np.linalg.norm(P3) ** 2 - np.linalg.norm(P4) ** 2) * 0.5
    R = (np.linalg.norm(P2) ** 2 - np.linalg.norm(P3) ** 2) * 0.5
    # 常数项系数
    L = np.ones((3, 1))
    L[0] = P
    L[1] = Q
    L[2] = R
    # 系数Dx
    Dx = np.ones((3, 3))
    Dx[:, :] = D[:, :]
    Dx[:, 0] = L[:, 0]
    detDx = np.linalg.det(Dx)
   
    # 系数Dy
    Dy = np.ones((3, 3))
    Dy[:, :] = D[:, :]
    Dy[:, 1] = L[:, 0]
    detDy = np.linalg.det(Dy)
    # 系数Dz
    Dz = np.ones((3, 3))
    Dz[:, :] = D[:, :]
    Dz[:, 2] = L[:, 0]
    detDz = np.linalg.det(Dz)
    # 计算球心
    xo = detDx / detD
    yo = detDy / detD
    zo = detDz / detD
    sphere_o = [xo, yo, zo]
    # 计算半径
    sphere_r = np.linalg.norm(P1 - sphere_o)

    return sphere_o, sphere_r

4. Calculation results

球心坐标:[-5.480253504546088, -0.08827109437986415, -0.01369803960562648]
球体半径:1.9874710684528454