5x^2+5y^2-9x-19y-26=0

解

5 x^{2} + 5 y^{2} - 9 x - 19 y - 26 = 0

(a, b) = (\frac{9}{10}, \frac{19}{10}), r = \frac{481}{5 2}

解答ステップ

5 x^{2} + 5 y^{2} - 9 x - 19 y - 26 = 0

5 x^{2} + 5 y^{2} - 9 x - 19 y - 26 = 0

を標準的な円のequationの形式で書き換える

(x - \frac{9}{10})^{2} + (y - \frac{19}{10})^{2} = (\frac{481}{5 2})^{2}

ゆえに円の特性は:

(a, b) = (\frac{9}{10}, \frac{19}{10}), r = \frac{481}{5 2}