数学 – 図形と数式の関係 - 平面図形と式 – 円と軌跡(アポロニウスの円)

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数学読本〈2〉簡単な関数/平面図形と式/指数関数・対数関数/三角関数 (松坂和夫(著)、岩波書店)の第6章(図形と数式の関係 - 平面図形と式)、6.3(円と軌跡)、軌跡の方程式、問29.を解いてみる。

問29.

$\begin{array}{l} P (x, y) \\ 2 A P = 3 P B \\ 4 A P^{2} = 9 P B^{2} \\ 4 ({(x + 3)}^{2} + y^{2}) = 9 ({(x - 2)}^{2} + y^{2}) \\ 4 (x^{2} + 6 x + 9 + y^{2}) = 9 (x^{2} - 4 x + 4 + y^{2}) \\ 5 x^{2} + 5 y^{2} - 60 x = 0 \\ {(x - 6)}^{2} + y^{2} = 36 \end{array}$
$\begin{array}{l} P (x, y) \\ n A P = m P B \\ n^{2} A P^{2} = m^{2} P B^{2} \\ n^{2} ({(x + m)}^{2} + y^{2}) = m^{2} ({(x - n)}^{2} + y^{2}) \\ (m^{2} - n^{2}) x^{2} + (m^{2} - n^{2}) y^{2} - 2 m^{2} n x - 2 m n^{2} x = 0 \\ (m^{2} - n^{2}) x^{2} + (m^{2} - n^{2}) y^{2} - 2 m n (m + n) x = 0 \\ (m - n) x^{2} + (m - n) y^{2} - 2 m n = 0 \\ {(x - \frac{m n}{m - n})}^{2} + y^{2} = {(\frac{m n}{m - n})}^{2} \end{array}$

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