144x^2+144y^2-96x+72y-11=0

解答

144 x^{2} + 144 y^{2} - 96 x + 72 y - 11 = 0

(a, b) = (\frac{1}{3}, - \frac{1}{4}), r = \frac{1}{2}

求解步骤

144 x^{2} + 144 y^{2} - 96 x + 72 y - 11 = 0

将

144 x^{2} + 144 y^{2} - 96 x + 72 y - 11 = 0

改写为圆标准方程

(x - \frac{1}{3})^{2} + (y - (- \frac{1}{4}))^{2} = (\frac{1}{2})^{2}

因此圆的性质是:

(a, b) = (\frac{1}{3}, - \frac{1}{4}), r = \frac{1}{2}

x^{2} + (y - 4)^{2} = 49

x^{2} + y^{2} - x - 10 y + 5 = 0

x^{2} + y^{2} + 5 x - 10 y - 5 = 0

a^{2} - 28 + b^{2} = 7

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{4} = 1