(x+4)^2+(y+2)^2=16

解答

(x + 4)^{2} + (y + 2)^{2} = 16

(a, b) = (- 4, - 2), r = 4

求解步骤

(x + 4)^{2} + (y + 2)^{2} = 16

将

(x + 4)^{2} + (y + 2)^{2} = 16

改写为圆标准方程

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

因此圆的性质是:

(a, b) = (- 4, - 2), r = 4

x^{2} + y^{2} - 4 x - 2 y = 11

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

x^{2} + 2 x + y^{2} = 3

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

(x - 2)^{2} + (y - 2)^{2} = 16