アポロニウスの円

平面上の2点からの距離の比が一定の点の軌跡は円になり，その円のことをアポロニウスの円という．

■証明1

平面上の2点を，点 $3$\text{O}$ と点 $3$\text{A}$ とする（図を参照）．さらに，点 $3$\text{P}$ があり，点 $3$\text{P}$ ，点 $3$\text{O}$ ，点 $3$\text{A}$ の関係は

$3$\displaystyle{\text{OP}:\text{AP}=m:n}$

となっている．

$3$\triangle \text{OAP}$ に関して

頂点 $3$\text{A}$ の内角の二等分線と辺 $3$\text{OA}$ の交点を $3$\hspace{1}{\text{X}}_{1}$ とすると，点 $3$\hspace{1}{\text{X}}_{1}$ は辺 $3$\text{OA}$ の内分点になる．

頂点 $3$\text{A}$ の外角の二等分線と辺 $3$\text{OA}$ の延長線との交点を $3$\hspace{1}{\text{X}}_{2}$ とすると，点 $3$\hspace{1}{\text{X}}_{2}$ は辺 $3$\text{OA}$ の外分点になる．

$3$\displaystyle{\angle \hspace{1}{\text{X}}_{1}\hspace{1}{\text{PX}}_{2}=90^\circ }$ となることより， $3$m:n$ が一定であれば，円周角の定理より

点 $3$\text{P}$ の軌跡は，線分 $3$\displaystyle{\hspace{1}{\text{X}}_{1}\hspace{1}{\text{X}}_{2}}$ を直径とする円

を描く

■証明2

座標平面を使って証明する．

２点の内１つは原点 $3$\text{O}$ とし，もう一方の点の座標を $3$\left({ a,0 }\right)$ とし点 $3$\text{A}$ とする．ただし， $3$\displaystyle{a> 0}$ とする．そして， $3$\text{OP}:\text{OA}=m:n$ となる点 $3$\text{P}$ の座標を $3$\displaystyle{\left({ x,y }\right)}$ とする．

上記内容を式で表すと

$3$\displaystyle{\sqrt{ {x}^{2}+{y}^{2} }:\sqrt{ { \left({ x- a }\right) }^{2}+{y}^{2} }=m:n}$ ･･････(1)

となる．この式を以下のように変形する．

$3$\displaystyle{n\sqrt{ {x}^{2}+{y}^{2} }=m\sqrt{ { \left({ x- a }\right) }^{2}+{y}^{2} }}$

$3$\displaystyle{{n}^{2}\left({ {x}^{2}+{y}^{2} }\right)={m}^{2}\left{{ { \left({ x- a }\right) }^{2}+{y}^{2} }\right}}$

$3$\displaystyle{{n}^{2}{x}^{2}+{n}^{2}{y}^{2}}$ $3$\displaystyle{={m}^{2}{x}^{2}- 2{m}^{2}ax+{m}^{2}{a}^{2}+{m}^{2}{y}^{2}}$

$3$\displaystyle{\left({ {n}^{2}- {m}^{2} }\right){x}^{2}+2{m}^{2}ax+\left({ {n}^{2}- {m}^{2} }\right){y}^{2}}$ $3$\displaystyle{={m}^{2}{a}^{2}}$

$3$\displaystyle{{x}^{2}+\frac{\hspace{2} 2{m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}x+{y}^{2}}$ $3$\displaystyle{=\frac{\hspace{2} {m}^{2}{a}^{2} \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}}$

$3$\displaystyle{{\left{{ x+\frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}} }\right}}^{2}+{y}^{2}}$ $3$\displaystyle{=\frac{\hspace{2} {m}^{2}{a}^{2} \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}+{\left{{ \frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}} }\right}}^{2}}$

$3$\displaystyle{{\left{{ x+\frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}} }\right}}^{2}+{y}^{2}}$ $3$\displaystyle{=\frac{\hspace{2} {m}^{2}{a}^{2}\left({ {n}^{2}- {m}^{2} }\right)+{m}^{4}{a}^{2} \hspace{2}}{\hspace{2} { \left({ {n}^{2}- {m}^{2} }\right) }^{2} \hspace{2}}}$

$3$\displaystyle{{\left{{ x+\frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}} }\right}}^{2}+{y}^{2}}$ $3$\displaystyle{=\frac{\hspace{2} {m}^{2}{n}^{2}{a}^{2} \hspace{2}}{\hspace{2} { \left({ {n}^{2}- {m}^{2} }\right) }^{2} \hspace{2}}}$

$3$\displaystyle{{\left{{ x+\frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}} }\right}}^{2}+{y}^{2}}$ $3$\displaystyle{={\left({ \frac{\hspace{2} mna \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}} }\right)}^{2}}$ ･･････(2)

(2)は円の方程式で，円の中心が， $3$\displaystyle{\left({ - \frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}},0 }\right)}$ ，半径が， $3$\displaystyle{\frac{\hspace{2} mna \hspace{2}}{\hspace{2} \left|{ {n}^{2}- {m}^{2} }\right| \hspace{2}}}$ の円を表す．

円の $3$x$ 軸との交点を求めてみる．(2)に $3$y=0$ を代入する．

$3$\displaystyle{{\left{{ x+\frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}} }\right}}^{2}={\left({ \frac{\hspace{2} mna \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}} }\right)}^{2}}$

$3$\displaystyle{{x}^{2}+\frac{\hspace{2} 2{m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}x+{\left({ \frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}} }\right)}^{2}- {\left({ \frac{\hspace{2} mna \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}} }\right)}^{2}=0}$

$3$\displaystyle{{x}^{2}+\frac{\hspace{2} 2{m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}x+\frac{\hspace{2} {m}^{4}{a}^{2}- {m}^{2}{n}^{2}{a}^{2} \hspace{2}}{\hspace{2} { \left({ {n}^{2}- {m}^{2} }\right) }^{2} \hspace{2}}=0}$

$3$\displaystyle{{x}^{2}+\frac{\hspace{2} 2{m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}x+\frac{\hspace{2} {m}^{2}{a}^{2}\left({ {m}^{2}- {n}^{2} }\right) \hspace{2}}{\hspace{2} { \left({ {n}^{2}- {m}^{2} }\right) }^{2} \hspace{2}}=0}$

$3$\displaystyle{{x}^{2}+\frac{\hspace{2} 2{m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}x- \frac{\hspace{2} {m}^{2}{a}^{2} \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}=0}$

解の公式より

$3$\displaystyle{x=- \frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}\pm \sqrt{ { \left{{ \frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} \left({ {n}^{2}- {m}^{2} }\right) \hspace{2}} }\right} }^{2}+\frac{\hspace{2} {m}^{2}{a}^{2} \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}} }}$

$3$\displaystyle{=- \frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}\pm \sqrt{ \frac{\hspace{2} {m}^{4}{a}^{2}+{m}^{2}{a}^{2}\left({ {n}^{2}- {m}^{2} }\right) \hspace{2}}{\hspace{2} { \left({ {n}^{2}- {m}^{2} }\right) }^{2} \hspace{2}} }}$

$3$\displaystyle{=- \frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}\pm \sqrt{ \frac{\hspace{2} {m}^{2}{n}^{2}{a}^{2} \hspace{2}}{\hspace{2} { \left({ {n}^{2}- {m}^{2} }\right) }^{2} \hspace{2}} }}$

$3$\displaystyle{=- \frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}\pm \frac{\hspace{2} mna \hspace{2}}{\hspace{2} \left|{ {n}^{2}- {m}^{2} }\right| \hspace{2}}}$ 　･･････(2)

$3$\displaystyle{\bullet }$ $3$\displaystyle{m< n}$ のとき，(2)は

$3$\displaystyle{x=- \frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}\pm \frac{\hspace{2} mna \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}}$ 　･･････(3)

$3$\displaystyle{\bullet }$ $3$\displaystyle{m> n}$ のとき，(2)は

$3$\displaystyle{x=- \frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}\pm \frac{\hspace{2} mna \hspace{2}}{\hspace{2} - \left({ {n}^{2}- {m}^{2} }\right) \hspace{2}}}$ $3$\displaystyle{=- \frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}\mp \frac{\hspace{2} mna \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}}$ 　･･････(4)

場合分けをしたが，結局，(3)と(4)は同じになる．

$3$\displaystyle{=- \frac{\hspace{2} {m}^{2}a \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}\pm \frac{\hspace{2} mna \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2}{ }^{2} \hspace{2}}}$

$3$\displaystyle{=- \frac{\hspace{2} {m}^{2}a\mp mna \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}}$

$3$\displaystyle{=- \frac{\hspace{2} ma\left({ m\mp n }\right) \hspace{2}}{\hspace{2} {n}^{2}- {m}^{2} \hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2} ma\left({ m\mp n }\right) \hspace{2}}{\hspace{2} {m}^{2}- {n}^{2} \hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2} ma\left({ m- n }\right) \hspace{2}}{\hspace{2} \left({ n- m }\right)\left({ n+m }\right) \hspace{2}}}$ ， $3$\displaystyle{\frac{\hspace{2} ma\left({ m+n }\right) \hspace{2}}{\hspace{2} \left({ n- m }\right)\left({ n+m }\right) \hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2} ma \hspace{2}}{\hspace{2} m+n \hspace{2}},\frac{\hspace{2} ma \hspace{2}}{\hspace{2} m- n \hspace{2}}}$

よって，円と $3$x$ 軸との交点の座標は

$3$\displaystyle{\left({ \frac{\hspace{2} ma \hspace{2}}{\hspace{2} m+n \hspace{2}},0 }\right)}$ ， $3$\displaystyle{\left({ \frac{\hspace{2} ma \hspace{2}}{\hspace{2} m- n \hspace{2}},0 }\right)}$

となる．

点 $3$\displaystyle{\left({ \frac{\hspace{2} ma \hspace{2}}{\hspace{2} m+n \hspace{2}},0 }\right)}$ は，線分 $3$\text{OA}$ を $3$\displaystyle{m:n}$ に内分した点であり，点 $3$\displaystyle{\left({ \frac{\hspace{2} ma \hspace{2}}{\hspace{2} m- n \hspace{2}},0 }\right)}$ は，線分 $3$\text{OA}$ を $3$\displaystyle{m:n}$ に外分した点である．

$3$m$ ， $3$n$ の値をスライダーの〇印をドラグして変更してみてください．また，点 $3$\text{P}$ をドラッグして動かしてみてください．

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最終更新日： 2025年11月25日