x^2+y^2+x+3y-104=0

解

x^{2} + y^{2} + x + 3 y - 104 = 0

(a, b) = (- \frac{1}{2}, - \frac{3}{2}), r = \frac{213}{2}

解答ステップ

x^{2} + y^{2} + x + 3 y - 104 = 0

x^{2} + y^{2} + x + 3 y - 104 = 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{3}{2}))^{2} = (\frac{213}{2})^{2}

ゆえに円の特性は:

(a, b) = (- \frac{1}{2}, - \frac{3}{2}), r = \frac{213}{2}