(x-4)^2+(y+1)^2=5

解答

(x - 4)^{2} + (y + 1)^{2} = 5

(a, b) = (4, - 1), r = 5

求解步骤

(x - 4)^{2} + (y + 1)^{2} = 5

将

(x - 4)^{2} + (y + 1)^{2} = 5

改写为圆标准方程

(x - 4)^{2} + (y - (- 1))^{2} = (5)^{2}

因此圆的性质是:

(a, b) = (4, - 1), r = 5

(x + 3)^{2} + (y - 7)^{2} = 9

(x - 3)^{2} + 2 y + 1 - 81 = - y^{2}

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

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

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