x^2+y^2+2(29/22)x+2(29/22)y+(-142/11)=0

解答

x^{2} + y^{2} + 2 (\frac{29}{22}) x + 2 (\frac{29}{22}) y + (- \frac{142}{11}) = 0

(a, b) = (- \frac{29}{22}, - \frac{29}{22}), r = \frac{3965}{11 2}

求解步骤

x^{2} + y^{2} + 2 \cdot \frac{29}{22} x + 2 \cdot \frac{29}{22} y + (- \frac{142}{11}) = 0

将

x^{2} + y^{2} + 2 \frac{29}{22} x + 2 \frac{29}{22} y + (- \frac{142}{11}) = 0

改写为圆标准方程

(x - (- \frac{29}{22}))^{2} + (y - (- \frac{29}{22}))^{2} = (\frac{3965}{11 2})^{2}

因此圆的性质是:

(a, b) = (- \frac{29}{22}, - \frac{29}{22}), r = \frac{3965}{11 2}

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

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