3x^2+3y^2-4x+y-12=0

解

3 x^{2} + 3 y^{2} - 4 x + y - 12 = 0

(a, b) = (\frac{2}{3}, - \frac{1}{6}), r = \frac{161}{6}

解答ステップ

3 x^{2} + 3 y^{2} - 4 x + y - 12 = 0

3 x^{2} + 3 y^{2} - 4 x + y - 12 = 0

を標準的な円のequationの形式で書き換える

(x - \frac{2}{3})^{2} + (y - (- \frac{1}{6}))^{2} = (\frac{161}{6})^{2}

ゆえに円の特性は:

(a, b) = (\frac{2}{3}, - \frac{1}{6}), r = \frac{161}{6}