(x-4)^2+(y-20)^2=sqrt(592)^2

解答

(x - 4)^{2} + (y - 20)^{2} = 592^{2}

(a, b) = (4, 20), r = (592)^{2}

求解步骤

(x - 4)^{2} + (y - 20)^{2} = (592)^{2}

将

(x - 4)^{2} + (y - 20)^{2} = 592^{2}

改写为圆标准方程

(x - 4)^{2} + (y - 20)^{2} = ((592)^{2})^{2}

因此圆的性质是:

(a, b) = (4, 20), r = (592)^{2}

x^{2} + x + y^{2} + \frac{8}{3} y - \frac{1223}{36} = 0

(x - 4)^{2} + (y - 2)^{2} = 36

x^{2} + y^{2} + 2 x - 10 y + 1 = 0

(x - 2)^{2} + (y + 8)^{2} = 144

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