3x^2+3y^2-3x-3y-9=0

解

3 x^{2} + 3 y^{2} - 3 x - 3 y - 9 = 0

(a, b) = (\frac{1}{2}, \frac{1}{2}), r = \frac{7}{2}

解答ステップ

3 x^{2} + 3 y^{2} - 3 x - 3 y - 9 = 0

3 x^{2} + 3 y^{2} - 3 x - 3 y - 9 = 0

を標準的な円のequationの形式で書き換える

(x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = (\frac{7}{2})^{2}

ゆえに円の特性は:

(a, b) = (\frac{1}{2}, \frac{1}{2}), r = \frac{7}{2}