(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

y^{2} + x^{2} - 2 x + 10 y = - 25

x^{2} + y^{2} - 10 x + 10 y + 25 = 0

(x - 5)^{2} + (y + 3)^{2} = 9

x^{2} + y^{2} + 4 x - 10 y + 4 = 0

(x - 5)^{2} + (y + 3)^{2} = 4