$Let the length of the intercepts on X - axis and Y - axis made by the circle$ $x^2+y^2+ax+2ay+c=0,$ $be$ $2\sqrt{2}$$and$$2\sqrt{5}$ $respectively. The dis\tan ce from origin to a \tan gent to this circle which is perpendicular to the line x+2y=0 is equal to$ 

$\begin{array}{l}\text{Length of }x-\text{\hspace{0.17em}intercept is 2}\sqrt{g^2-c}=2\sqrt{\frac{a^2}{4}-c}=2\sqrt{2}\Rightarrow a^2-4c=8\\ \text{Length of }y-\text{\hspace{0.17em}intercept is 2}\sqrt{f^2-c}=2\sqrt{a^2-c}=2\sqrt{5}\Rightarrow a^2-c=5\\ \text{From the above two eqautions }3c=-3\Rightarrow c=-1\\ a^2=4\Rightarrow a=-2,a<0\\ \text{equation of the circle is }{x}^2+y^2-2x-4y-1=0,\text{center }\left(1,2\right)\text{\hspace{0.17em}radius is }\sqrt{1+4+1}=\sqrt{6}\\ \text{Equation of any line perpendicular to the line x+2y=0 is 2x-y+k=0}\\ This is \tan gent to the above circle, \\ Hence, \sqrt{6}=\frac{2-2+k}{\sqrt{5}}\Rightarrow k=\sqrt{30}\\ Therefore, the equation of \tan gent is 2x-y+\sqrt{30}=0\\ The perpendicular dis\tan ce from origin to the above line is \frac{\sqrt{30}}{\sqrt{5}}=\sqrt{6}\end{array}$