5x^2+5y^2+10x-5y+3=0

解

5 x^{2} + 5 y^{2} + 10 x - 5 y + 3 = 0

(a, b) = (- 1, \frac{1}{2}), r = \frac{13}{2 5}

解答ステップ

5 x^{2} + 5 y^{2} + 10 x - 5 y + 3 = 0

5 x^{2} + 5 y^{2} + 10 x - 5 y + 3 = 0

を標準的な円のequationの形式で書き換える

(x - (- 1))^{2} + (y - \frac{1}{2})^{2} = (\frac{13}{2 5})^{2}

ゆえに円の特性は:

(a, b) = (- 1, \frac{1}{2}), r = \frac{13}{2 5}