2x^2+2y^2-3x+5y-4=0

解

2 x^{2} + 2 y^{2} - 3 x + 5 y - 4 = 0

(a, b) = (\frac{3}{4}, - \frac{5}{4}), r = \frac{33}{2 2}

解答ステップ

2 x^{2} + 2 y^{2} - 3 x + 5 y - 4 = 0

2 x^{2} + 2 y^{2} - 3 x + 5 y - 4 = 0

を標準的な円のequationの形式で書き換える

(x - \frac{3}{4})^{2} + (y - (- \frac{5}{4}))^{2} = (\frac{33}{2 2})^{2}

ゆえに円の特性は:

(a, b) = (\frac{3}{4}, - \frac{5}{4}), r = \frac{33}{2 2}