数学 – 図形と数式の関係 - 平面図形と式 – 円と軌跡(点の軌跡を求める)

学習環境

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30.を解いてみる。

問30.

$\begin{array}{l} P (x, y) \\ A P = 2 P B \\ A P^{2} = 4 P B^{2} \\ {(x - 6)}^{2} + y^{2} = 4 (x^{2} + {(y - 3)}^{2}) \\ 3 x^{2} + 3 y^{2} + 12 x - 24 y = 0 \\ x^{2} + y^{2} + 4 x - 8 y = 0 \\ {(x + 2)}^{2} + {(y - 4)}^{2} = 20 \end{array}$
$\begin{array}{l} P (x, y) \\ ({(x + 1)}^{2} + y^{2}) + ({(x - 1)}^{2} + y^{2}) = 4 \\ 2 x^{2} + 2 y^{2} = 2 \\ x^{2} + y^{2} = 1 \end{array}$
$\begin{array}{l} P (x, y) \\ x^{2} + y^{2} = ({(x - 1)}^{2} + y^{2}) + ({(x - 2)}^{2} + y^{2}) \\ x^{2} + y^{2} - 6 x + 5 = 0 \\ {(x - 3)}^{2} + y^{2} = 4 \end{array}$
$\begin{array}{l} P (x, y) \\ (x^{2} + y^{2}) + ({(x - 4)}^{2} + {(y - 2)}^{2}) + ({(x - 5)}^{2} + {(y - 1)}^{2}) = 64 \\ 3 x^{2} + 3 y^{2} - 18 x - 6 y - 18 = 0 \\ x^{2} + y^{2} - 9 x - 2 y - 6 = 0 \\ {(x - 3)}^{2} + {(y - 1)}^{2} = 16 \end{array}$
$\begin{array}{l} P (x, y) \\ Q (u, v) \\ x = \frac{3 u}{3 + 1} \\ y = \frac{3 v}{3 + 1} \\ u = \frac{4}{3} x \\ v = \frac{4}{3} y \\ u^{2} + v^{2} - 6 u - 10 v + 30 = 0 \\ \frac{16}{9} x^{2} + \frac{16}{9} y^{2} - 8 x - \frac{40}{3} y + 30 = 0 \\ x^{2} + y^{2} - \frac{9}{2} x - \frac{15}{2} y + \frac{270}{16} = 0 \\ {(x - \frac{9}{4})}^{2} + {(y - \frac{15}{4})}^{2} + \frac{270}{16} - \frac{81}{16} - \frac{225}{16} = 0 \\ {(x - \frac{9}{4})}^{2} + {(y - \frac{15}{4})}^{2} = \frac{9}{4} \end{array}$

Kamimura's blog