Let the lengths of intercepts on the x-axis a | vedclass

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(C) The equation of the circle is $x^{2}+y^{2}+ax+2ay+c=0$.The length of the $x$-intercept is $2\sqrt{g^{2}-c} = 2\sqrt{\frac{a^{2}}{4}-c} = 2\sqrt{2}

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$\sqrt{6}$

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(C) The equation of the circle is $x^{2}+y^{2}+ax+2ay+c=0$.The length of the $x$-intercept is $2\sqrt{g^{2}-c} = 2\sqrt{\frac{a^{2}}{4}-c} = 2\sqrt{2}$.$\Rightarrow \frac{a^{2}}{4}-c = 2 \quad \dots(1)$The length of the $y$-intercept is $2\sqrt{f^{2}-c} = 2\sqrt{a^{2}-c} = 2\sqrt{5}$.$\Rightarrow a^{2}-c = 5 \quad \dots(2)$Subtracting $(1)$ from $(2)$:$(a^{2}-c) - (\frac{a^{2}}{4}-c) = 5-2$ $\Rightarrow \frac{3a^{2}}{4} = 3$ $\Rightarrow a^{2} = 4$.Since $a Substituting $a = -2$ into $(2)$: $(-2)^{2}-c = 5$ $\Rightarrow 4-c = 5$ $\Rightarrow c = -1$.The circle equation is $x^{2}+y^{2}-2x-4y-1 = 0$,which is $(x-1)^{2}+(y-2)^{2} = 6$.The center is $(1, 2)$ and the radius $r = \sqrt{6}$.The tangent is perpendicular to $x+2y=0$,so its slope $m = 2$.The equation of the tangent is $(y-2) = 2(x-1) \pm \sqrt{6}\sqrt{1+2^{2}}$.$y-2 = 2x-2 \pm \sqrt{30} \Rightarrow 2x-y \pm \sqrt{30} = 0$.The distance from the origin $(0,0)$ to the tangent $2x-y \pm \sqrt{30} = 0$ is $d = \frac{|\pm \sqrt{30}|}{\sqrt{2^{2}+(-1)^{2}}} = \frac{\sqrt{30}}{\sqrt{5}} = \sqrt{6}$.

Let the lengths of intercepts on the $x$-axis and $y$-axis made by the circle $x^{2}+y^{2}+ax+2ay+c=0$ $(a < 0)$ be $2\sqrt{2}$ and $2\sqrt{5}$,respectively. Then the shortest distance from the origin to a tangent to this circle which is perpendicular to the line $x+2y=0$ is equal to:

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