原問題
https://math.pro/db/viewthread.php?tid=3490&page=1#pid22082
使用maxima計算結果如下
設半徑4的球心\(A(0,0,0)\)，半徑9的球心\(B(13,0,0)\)，半徑16的球心\(\displaystyle C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)\)
外公切平面方程式為\(30\sqrt{29}x+271y\pm 13\sqrt{455}z+312\sqrt{29}=0\)
和半徑4的球相切於\(\displaystyle A' \left(-\frac{20}{13},-\frac{542}{39\sqrt{29}},-\frac{2\sqrt{455}}{3\sqrt{29}}\right)\)
和半徑9的球相切於\(\displaystyle B' \left(\frac{124}{13},-\frac{813}{26\sqrt{29}},-\frac{3\sqrt{455}}{2\sqrt{29}}\right)\)
和半徑16的球相切於\(\displaystyle C' \left(-\frac{108}{13},\frac{2008}{39\sqrt{29}},-\frac{8\sqrt{455}}{3\sqrt{29}}\right)\)
有前一次的經驗後就直接計算各點位置
設半徑4的球心\(A(0,0,0)\)，半徑9的球心\(B(13,0,0)\)，半徑16的球心\(C(x,y,0)\)
\(\cases{\overline{BC}=25\cr \overline{CA}=20}\)，\(\cases{(x-13)^2+(y-0)^2=25^2 \cr (x-0)^2+(y-0)^2=20^2}\)，
\(\displaystyle x=-\frac{28}{13},y=\frac{48\sqrt{29}}{13}\)，得到\(\displaystyle C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)\)
設\(P(x,y,0)\)為兩個外公切平面方程式交線上一點
球心\(A(0,0,0)\)半徑4的球面和平面相切於\(A'\)點，\(\Rightarrow \overline{AA'}=4\)，\(∠PA'A=90^{\circ}\)
球心\(B(13,0,0)\)半徑9的球面和平面相切於\(B'\)點，\(\Rightarrow \overline{BB'}=9\)，\(∠PB'B=90^{\circ}\)
\(\Delta PAA'\)和\(\Delta PBB'\)為相似三角形(\(∠APA'=∠BPB'\)，\(∠PA'A=∠PB'B=90^{\circ}\))
\(\overline{PA}:\overline{PB}=\overline{AA'}:\overline{BB'}=4:9\)，\(\overline{PA}:\overline{AB}=4:5\)
由外分點公式可知\(\displaystyle P=\frac{9}{5}A-\frac{4}{5}B=\frac{9}{5}(0,0,0)-\frac{4}{5}(13,0,0)=(-\frac{52}{5},0,0)\)
得到\(\displaystyle P(-\frac{52}{5},0,0)\)
設\(Q(x,y,0)\)為兩個外公切平面方程式交線上一點
球心\(A(0,0,0)\)半徑4的球面和平面相切於\(A'\)點，\(\Rightarrow \overline{AA'}=4\)，\(∠PA'A=90^{\circ}\)
球心\(\displaystyle C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)\)半徑16的球面和平面相切於\(C'\)點，\(\Rightarrow \overline{CC'}=16\)，\(∠PC'C=90^{\circ}\)
\(\Delta PAA'\)和\(\Delta PCC'\)為相似三角形(\(∠APA'=∠CPC'\)，\(∠PA'A=∠PC'C=90^{\circ}\))
\(\overline{QA}:\overline{QC}=\overline{AA'}:\overline{CC'}=4:16\)，\(\overline{QA}:\overline{AC}=1:3\)
由外分點公式可知\(\displaystyle Q=\frac{4}{3}A-\frac{1}{3}C=\frac{4}{3}(0,0,0)-\frac{1}{3}C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)=\left(\frac{28}{39},-\frac{16\sqrt{29}}{13},0\right)\)
得到\(\displaystyle Q\left(\frac{28}{39},-\frac{16\sqrt{29}}{13},0\right)\)
從\(\displaystyle P\left(-\frac{52}{5},0,0\right)\)和\(\displaystyle Q\left(\frac{28}{39},-\frac{16\sqrt{29}}{13},0\right)\)求交線的對稱比例式
方向向量為\(\displaystyle \vec{PQ}=\left(\frac{2168}{195},-\frac{16\sqrt{29}}{13},0\right)\)
\(PQ\)直線的對稱比例式為\(\displaystyle \frac{x+\frac{52}{5}}{\frac{2168}{195}}=\frac{y-0}{-\frac{16\sqrt{29}}{13}}\)，\(z=0\)
化簡得到\(30\sqrt{29}x+271y+312\sqrt{29}=0\)，\(z=0\)
假設外公切平面方程式為\(30\sqrt{29}x+271y+312\sqrt{29}+kz=0\)，\(k\in R\)
球心\(A(0,0,0)\)到平面距離\(\displaystyle \frac{|\;0+0+0+312\sqrt{29}|\;}{\sqrt{(30\sqrt{29})^2+271^2+k^2}}=4\)，\(k=\pm 13\sqrt{455}\)
得到外公切平面方程式\(30\sqrt{29}x+271y\pm 13\sqrt{455}z+312\sqrt{29}=0\)
利用投影點公式求切點\(A',B',C'\)
投影點公式：
\(P(x_0,y_0,z_0)\)對平面\(ax+by+cz+d=0\)的投影點為\(P'(x_0-at,y_0-bt,z_0-ct)\)，其中\(\displaystyle t=\frac{ax_0+by_0+cz_0+d}{a^2+b^2+c^2}\)
計算\(A(0,0,0)\)的投影點\(A'\)
\(\displaystyle t=\frac{0+0+0+312\sqrt{29}}{(30\sqrt{29})^2+271^2+(13\sqrt{455})^2}=\frac{2}{39\sqrt{29}}\)
\(\displaystyle A'\left(0-30\sqrt{29}\cdot \frac{2}{39\sqrt{29}},0-271\cdot \frac{2}{39\sqrt{29}},0-13\sqrt{455}\cdot \frac{2}{39\sqrt{29}}\right)=
\left(-\frac{30}{13},-\frac{542}{39\sqrt{29}},-\frac{2\sqrt{455}}{3\sqrt{29}}\right)\)
計算\(B(13,0,0)\)的投影點\(B'\)
\(\displaystyle t=\frac{30\sqrt{29}\cdot 13+0+0+312\sqrt{29}}{(30\sqrt{29})^2+271^2+(13\sqrt{455})^2}=\frac{3}{26\sqrt{29}}\)
\(\displaystyle B'\left(13-30\sqrt{29}\cdot \frac{3}{26\sqrt{29}},0-271\cdot \frac{3}{26\sqrt{29}},0-13\sqrt{455}\cdot \frac{3}{26\sqrt{29}}\right)=
\left(\frac{124}{13},-\frac{813}{26\sqrt{29}},-\frac{3\sqrt{455}}{2\sqrt{29}}\right)\)
計算\(\displaystyle C\left(-\frac{28}{13},\frac{48\sqrt{29}}{13},0\right)\)的投影點\(C'\)
\(\displaystyle t=\frac{30\sqrt{29}\cdot (-\frac{28}{13})+271\cdot \frac{48\sqrt{29}}{13}+0+312\sqrt{29}}{(30\sqrt{29})^2+271^2+(13\sqrt{455})^2}=\frac{8}{39\sqrt{29}}\)
\(\displaystyle C'\left(-\frac{28}{13}-30\sqrt{29}\cdot \frac{8}{39\sqrt{29}},\frac{48\sqrt{29}}{13}-271\cdot \frac{8}{39\sqrt{29}},0-13\sqrt{455}\cdot \frac{8}{39\sqrt{29}}\right)=
\left(-\frac{108}{13},\frac{2008}{39\sqrt{29}},-\frac{8\sqrt{455}}{3\sqrt{29}}\right)\)
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