2x^2+2y^2-16x+4y+20=0

解答

2 x^{2} + 2 y^{2} - 16 x + 4 y + 20 = 0

(a, b) = (4, - 1), r = 7

求解步骤

2 x^{2} + 2 y^{2} - 16 x + 4 y + 20 = 0

将

2 x^{2} + 2 y^{2} - 16 x + 4 y + 20 = 0

改写为圆标准方程

(x - 4)^{2} + (y - (- 1))^{2} = (7)^{2}

因此圆的性质是:

(a, b) = (4, - 1), r = 7

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

x^{2} + y^{2} + 2 x - 8 y - 19 = 0

(x + 2)^{2} + (y + 2)^{2} = 9

x^{2} + y^{2} + 8 x - 2 y + 15 = 0

(x + 1)^{2} + y^{2} = 36