问题及解答

已知四点 $A=(1,2,7)$, $B=(4,3,3)$, $C=(5,-1,6)$, $D=(\sqrt{7},\sqrt{7},0)$. 试求过这四点的球面方程.

设所求球面的圆心为 $O=(x,y,z)$, 半径为 $r$. 则球面方程为

\[
(X-x)^2+(Y-y)^2+(Z-z)^2=r^2.
\]

将 $A,B,C,D$ 四点的坐标代入, 得

\[
\left\{
\begin{aligned}
(x-1)^2+(y-2)^2+(z-7)^2=r^2,\\
(x-4)^2+(y-3)^2+(z-3)^2=r^2,\\
(x-5)^2+(y+1)^2+(z-6)^2=r^2,\\
(x-\sqrt{7})^2+(y-\sqrt{7})^2+z^2=r^2,\\
\end{aligned}
\right.
\]

整理得

\[
\left\{
\begin{aligned}
(x^2-2x+1)+(y^2-4y+4)+(z^2-14z+49)=r^2,\qquad (1)\\
(x^2-8x+16)+(y^2-6y+9)+(z^2-6z+9)=r^2,\qquad (2)\\
(x^2-10x+25)+(y^2+2y+1)+(z^2-12z+36)=r^2,\qquad (3)\\
(x^2-2\sqrt{7}x+7)+(y^2-2\sqrt{7}y+7)+z^2=r^2.\qquad (4)\\
\end{aligned}
\right.
\]

(2)-(1), (3)-(1), (4)-(1) 得到

\[
\left\{
\begin{aligned}
-6x-2y+8z-20=0,\\
-8x+6y+2z+8=0,\\
(2-2\sqrt{7})x+(4-2\sqrt{7})y+14z-40=0.\\
\end{aligned}
\right.
\]

化简得方程组

\[
\left\{
\begin{aligned}
3x+y-4z&=-10,\\
​4x-3y-z&=4,\\
​(1-\sqrt{7})x+(2-\sqrt{7})y+7z&=20.
\end{aligned}
\right.
\]

对其增广矩阵进行初等行变换

\[
\begin{split}
(A|b)&=\left(\begin{matrix}
3 & 1 & -4 \\
4 & -3 & -1\\
1-\sqrt{7} & 2-\sqrt{7} & 7\\
\end{matrix}\right|\left.
\begin{matrix}
-10\\
4\\
20\\
\end{matrix}\right)\\
&\xrightarrow{r_1\times\frac{1}{3}}
\left(\begin{matrix}
1 & \frac{1}{3} & -\frac{4}{3} \\
4 & -3 & -1\\
1-\sqrt{7} & 2-\sqrt{7} & 7\\
\end{matrix}\right|\left.
\begin{matrix}
-\frac{10}{3}\\
4\\
20\\
\end{matrix}\right)\\
&\xrightarrow[r_3-(1-\sqrt{7})r_1]{r_2-4r_1}
\left(\begin{matrix}
1 & \frac{1}{3} & -\frac{4}{3} \\
0 & -\frac{13}{3} & \frac{13}{3}\\
0 & \frac{5-2\sqrt{7}}{3} & \frac{25-4\sqrt{7}}{3}\\
\end{matrix}\right|\left.
\begin{matrix}
-\frac{10}{3}\\
4\cdot\frac{13}{3}\\
\frac{70-10\sqrt{7}}{3}\\
\end{matrix}\right)\\
&\xrightarrow{}
\left(\begin{matrix}
1 & \frac{1}{3} & -\frac{4}{3} \\
0 & 1 & -1\\
0 & 5-2\sqrt{7} & 25-4\sqrt{7}\\
\end{matrix}\right|\left.
\begin{matrix}
-\frac{10}{3}\\
-4\\
70-10\sqrt{7}\\
\end{matrix}\right)\\
&\xrightarrow{r_3-(5-2\sqrt{7})r_2}
\left(\begin{matrix}
1 & \frac{1}{3} & -\frac{4}{3} \\
0 & 1 & -1\\
0 & 0 & 30-6\sqrt{7}\\
\end{matrix}\right|\left.
\begin{matrix}
-\frac{10}{3}\\
-4\\
90-18\sqrt{7}\\
\end{matrix}\right)\\
&\xrightarrow{}
\left(\begin{matrix}
1 & \frac{1}{3} & -\frac{4}{3} \\
0 & 1 & -1\\
0 & 0 & 1\\
\end{matrix}\right|\left.
\begin{matrix}
-\frac{10}{3}\\
-4\\
3\\
\end{matrix}\right)\\
&\xrightarrow[r_1+\frac{4}{3}r_3]{r_2+r_3}
\left(\begin{matrix}
1 & \frac{1}{3} & 0 \\
0 & 1 & 0\\
0 & 0 & 1\\
\end{matrix}\right|\left.
\begin{matrix}
\frac{2}{3}\\
-1\\
3\\
\end{matrix}\right)\\
&\xrightarrow{r_1-\frac{1}{3}r_2}
\left(\begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0\\
0 & 0 & 1\\
\end{matrix}\right|\left.
\begin{matrix}
1\\
-1\\
3\\
\end{matrix}\right)\\
\end{split}
\]

因此圆心为 $(x,y,z)=(1,-1,3)$. 半径满足 $r^2=(1-1)^2+(-1-2)^2+(3-7)^2=25$, 所以半径为 $r=5$. 因此球面方程为

\[
(x-1)^2+(y+1)^2+(z-3)^2=25.
\]