中心 (x-7/4)^2+(y+3/4)^2= 169/16

解答

中心

(x - \frac{7}{4})^{2} + (y + \frac{3}{4})^{2} = \frac{169}{16}

(\frac{7}{4}, - \frac{3}{4})

求解步骤

(x - \frac{7}{4})^{2} + (y + \frac{3}{4})^{2} = \frac{169}{16}

将

(x - \frac{7}{4})^{2} + (y + \frac{3}{4})^{2} = \frac{169}{16}

改写为圆标准方程

(x - \frac{7}{4})^{2} + (y - (- \frac{3}{4}))^{2} = (\frac{13}{4})^{2}

因此圆的性质是:

(a, b) = (\frac{7}{4}, - \frac{3}{4}), r = \frac{13}{4}

中心是:

(\frac{7}{4}, - \frac{3}{4})

ce n t er 4 x^{2} + 4 y^{2} - 16 x - 24 y + 51 = 0

ce n t er 2 x^{2} + 2 y^{2} + 10 x - 6 y - 55 = 0

ce n t er x^{2} + y^{2} - 12 x - 10 y = - 12

ce n t er x^{2} + y^{2} - 4 x = 5

ce n t er 2 x^{2} - 6 x + 2 y^{2} + 2 y = 45