x^2+9x+y^2-5y-7=0

解答

x^{2} + 9 x + y^{2} - 5 y - 7 = 0

(a, b) = (- \frac{9}{2}, \frac{5}{2}), r = \frac{67}{2}

求解步骤

x^{2} + 9 x + y^{2} - 5 y - 7 = 0

将

x^{2} + 9 x + y^{2} - 5 y - 7 = 0

改写为圆标准方程

(x - (- \frac{9}{2}))^{2} + (y - \frac{5}{2})^{2} = (\frac{67}{2})^{2}

因此圆的性质是:

(a, b) = (- \frac{9}{2}, \frac{5}{2}), r = \frac{67}{2}

x^{2} + y^{2} - 2 x = 1

(x - 4)^{2} + (y - 5)^{2} = 4

(x - 4)^{2} + (y - 5)^{2} = 2

100 x^{2} + 100 y^{2} - 100 x + 240 y - 56 = 0

x^{2} + y^{2} + 3 y + \frac{7}{4} = 0