【数学A】例題3.2.7：共通接線（One More）★★★

Name: 【数学A】例題3.2.7：共通接線（One More）★★★
Uploaded: 2025-04-11T14:03:28+00:00

2025年4月11日2026年8月5日

問題の解答

検索用コード（LaTeX）

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右の図のように，半径6の円Oと半径10の円$\mathrm{O}^{\prime}$があり，中心間の距離$\mathrm{OO}^{\prime}=20$とする．2つの円の共通接線を2本引き，これらの接点をA，B，C，Dとするとき，線分AB，CDの長さを求めよ．

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Oから$\mathrm{O}^{\prime}$Bに垂線OHを下ろすと，$\angle\mathrm{OAB}=\angle\mathrm{O^\prime BA}=90^\circ$であるから，

\[
\mathrm{AB}=\mathrm{OH},\mathrm{BH}=\mathrm{AO}=6
\]

$\triangle\mathrm{OO}^{\prime}\mathrm{H}$において，$\angle\mathrm{OHO^\prime}=90^{\circ}$であるから，

\begin{align*}
\mathrm{OH}^2&=\mathrm{OO}^{\prime 2}-\mathrm{O}^{\prime}\mathrm{H}^2\\
&=20^2-(10-6)^2\\
&=20^2-4^2\\
&=384
\end{align*}

$\mathrm{OH}>0$より，$\mathrm{OH}=\sqrt{384}=8\sqrt{6}$

よって，$\mathrm{AB}=\mathrm{OH}=8\sqrt{6}$

Oから線分$\mathrm{O}^{\prime}\mathrm{D}$の延長に垂線$\mathrm{OH}^{\prime}$を下ろすと，$\angle\mathrm{OCD}=\angle\mathrm{O^\prime DC}=90^{\circ}$

したがって，$\mathrm{CD}=\mathrm{OH}^{\prime},\mathrm{DH}^{\prime}=\mathrm{CO}=6$

$\triangle\mathrm{OO}^{\prime}\mathrm{H}^{\prime}$において，$\angle\mathrm{OH^{\prime}O^\prime}=90^{\circ}$であるから，

\begin{align*}
\mathrm{OH}^{\prime 2}&=\mathrm{OO}^{\prime 2}-\mathrm{O}^{\prime}\mathrm{H}^{\prime 2}\\
&=20^2-(10+6)^2\\
&=20^2-16^2\\
&=144
\end{align*}

$\mathrm{OH}^{\prime}>0$より，$\mathrm{OH}^{\prime}=\sqrt{144}=12$

よって，$\mathrm{CD}=\mathrm{OH}^{\prime}=12$

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右の図のように，半径8の円Oと半径4の円$\mathrm{O}^{\prime}$があり，中心間の距離$\mathrm{OO}^{\prime}=13$とする．2つの円の共通接線を2本引き，これらの接点をA，B，C，Dとするとき，線分AB，CDの長さを求めよ．

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% 問題A3.2.7の解答（One More）★★★
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$\mathrm{O}^{\prime}$からOAに垂線$\mathrm{O}^{\prime}\mathrm{H}$を下ろすと，$\angle\mathrm{OAB}=\angle\mathrm{O^\prime BA}=90^\circ$であるから，

\[
\mathrm{AB}=\mathrm{O}^{\prime}\mathrm{H},\mathrm{AH}=\mathrm{B}\mathrm{O}^{\prime}=4
\]

$\triangle\mathrm{OO}^{\prime}\mathrm{H}$において，$\angle\mathrm{OHO^\prime}=90^{\circ}$であるから，

\begin{align*}
\mathrm{O}^{\prime}\mathrm{H}^2&=\mathrm{OO}^{\prime 2}-\mathrm{O}\mathrm{H}^2\\
&=13^2-(8-4)^2\\
&=13^2-4^2\\
&=153
\end{align*}

$\mathrm{O}^{\prime}\mathrm{H}>0$より，$\mathrm{O}^{\prime}\mathrm{H}=\sqrt{153}=3\sqrt{17}$

よって，$\mathrm{AB}=\mathrm{O}^{\prime}\mathrm{H}=3\sqrt{17}$

Oから線分$\mathrm{O}^{\prime}\mathrm{D}$の延長に垂線$\mathrm{OH}^{\prime}$を下ろすと，$\angle\mathrm{OCD}=\angle\mathrm{O^\prime DC}=90^{\circ}$

したがって，$\mathrm{CD}=\mathrm{OH}^{\prime},\mathrm{DH}^{\prime}=\mathrm{CO}=8$

$\triangle\mathrm{OO}^{\prime}\mathrm{H}^{\prime}$において，$\angle\mathrm{OH^{\prime}O^\prime}=90^{\circ}$であるから，

\begin{align*}
\mathrm{OH}^{\prime 2}&=\mathrm{OO}^{\prime 2}-\mathrm{O}^{\prime}\mathrm{H}^{\prime 2}\\
&=13^2-(4+8)^2\\
&=13^2-12^2\\
&=169-144\\
&=25
\end{align*}

$\mathrm{OH}^{\prime}>0$より，$\mathrm{OH}^{\prime}=\sqrt{25}=5$

よって，$\mathrm{CD}=\mathrm{OH}^{\prime}=5$

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% 例題A3.2.7の側注：共通接線（One More）★★★
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% 側注1
三平方の定理を用いる．

% 側注2
三平方の定理を用いる．

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% 問題A3.2.7の側注：共通接線（One More）★★★
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% 側注1
三平方の定理を用いる．

% 側注2
三平方の定理を用いる．

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% 例題A3.2.7の考え方・補足など：共通接線（One More）★★★
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補助線を引くと，直角三角形を見つけることができる．三平方の定理を用いて，線分の長さを求める．