3x^2+3y^2-5x+2y+1=0

解

3 x^{2} + 3 y^{2} - 5 x + 2 y + 1 = 0

(a, b) = (\frac{5}{6}, - \frac{1}{3}), r = \frac{17}{6}

解答ステップ

3 x^{2} + 3 y^{2} - 5 x + 2 y + 1 = 0

3 x^{2} + 3 y^{2} - 5 x + 2 y + 1 = 0

を標準的な円のequationの形式で書き換える

(x - \frac{5}{6})^{2} + (y - (- \frac{1}{3}))^{2} = (\frac{17}{6})^{2}

ゆえに円の特性は:

(a, b) = (\frac{5}{6}, - \frac{1}{3}), r = \frac{17}{6}