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

$$\begin{gathered} \begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{ax}+2\mathrm{ay}+\mathrm{c}=0\\ 2\sqrt{{\mathrm{g}}^2-\mathrm{c}}=2\sqrt{\frac{{\mathrm{a}}^2}{4}-\mathrm{c}}=2\sqrt{2}\\ \Rightarrow \frac{{\mathrm{a}}^2}{4}-\mathrm{c}=2 ...\left(1\right)\\ 2\sqrt{{\mathrm{f}}^2-\mathrm{c}}=2\sqrt{{\mathrm{a}}^2-\mathrm{c}}=2\sqrt{5}\\ \Rightarrow {\mathrm{a}}^2-\mathrm{c}=5 ...\left(2\right) \\ \left(1\right) \& \left(2\right) \\ \begin{array}{l}\frac{3{\mathrm{a}}^2}{4}=3\Rightarrow \mathrm{a}=-2(\mathrm{a}<0)\\ \therefore \mathrm{c}=-1 \\ \mathrm{Equation} \mathrm{of} \mathrm{Circle} \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-4\mathrm{y}-1=0 \\ \mathrm{Given} \mathrm{x}+2\mathrm{y}=0\Rightarrow \mathrm{m}=-\frac{1}{2} \\ \begin{array}{l}{\mathrm{m}}_{\text{tangent }}=2\\ \text{ Equation of tangent }\\ \Rightarrow (\mathrm{y}-2)=2(\mathrm{x}-1)\pm \sqrt{6}\sqrt{1+4}\\ \Rightarrow 2\mathrm{x}-\mathrm{y}\pm \sqrt{30}=0\end{array}\end{array}\end{array} \end{gathered}$$
Perpendicular distance from $(0,0)=\left|\frac{\pm \sqrt{30}}{\sqrt{4+1}}\right|=\sqrt{6}$