(x-7)^2+(y+3)^2-81=0

解答

(x - 7)^{2} + (y + 3)^{2} - 81 = 0

(a, b) = (7, - 3), r = 9

求解步骤

(x - 7)^{2} + (y + 3)^{2} - 81 = 0

将

(x - 7)^{2} + (y + 3)^{2} - 81 = 0

改写为圆标准方程

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

因此圆的性质是:

(a, b) = (7, - 3), r = 9

3 x^{2} + 3 y^{2} = 6

2 x^{2} + 2 y^{2} - 6 x - 14 y - 3 = 0

4 x^{2} + 4 y^{2} - 16 x + 12 y + 21 = 0

2 x^{2} + 3 x + 2 y^{2} + 7 y = 57.75

x^{2} + y^{2} - 2 x + 6 y - 90 = 0