Given circle

Use intercept on x-axis

On the $x$-axis, $y=0$. Then circle becomes $x^2+ax+c=0.$

The intercept length on $x$-axis is the distance between the two roots of this quadratic. For $x^2+ax+c=0$, root difference has magnitude $|x_1-x_2|=\sqrt{a^2-4c}.$

Given intercept length is $2\sqrt2$, so $\sqrt{a^2-4c}=2\sqrt2$ $a^2-4c=8. \qquad (1)$

Use intercept on y-axis

On the $y$-axis, $x=0$. Then circle becomes $y^2+2ay+c=0.$

Distance between its roots is $|y_1-y_2|=\sqrt{(2a)^2-4c}=\sqrt{4a^2-4c}=2\sqrt{a^2-c}.$

Given intercept length is $2\sqrt5$, so $2\sqrt{a^2-c}=2\sqrt5$ $a^2-c=5. \qquad (2)$

Find $a$ and $c$

From (2), $c=a^2-5.$

Substitute into (1): $a^2-4(a^2-5)=8$ $a^2-4a^2+20=8$ $-3a^2=-12$ $a^2=4.$

Since $a<0$, we get $a=-2.$

Then $c=a^2-5=4-5=-1.$

Find center and radius

Center: $\left(-\frac a2,-a\right)=\left(1,2\right).$

Radius: $r^2=\frac{5a^2}{4}-c=\frac{5\cdot 4}{4}-(-1)=5+1=6$ $r=\sqrt6.$

Tangents perpendicular to $x+2y=0$

The line $x+2y=0$ has slope $-\frac12$. Therefore a perpendicular tangent has slope $2$.

So tangent family is $y=2x+k$ or $2x-y+k=0.$

Condition for tangency

Distance from center $(1,2)$ to tangent $2x-y+k=0$ must equal radius $\sqrt6$: $\frac{|2(1)-2+k|}{\sqrt{2^2+(-1)^2}}=\sqrt6$ $\frac{|k|}{\sqrt5}=\sqrt6$ $|k|=\sqrt{30}.$

So the two tangents are $2x-y\pm\sqrt{30}=0.$

Shortest distance from origin to these tangents

Distance from origin $(0,0)$ to line $2x-y+k=0$ is $\frac{|k|}{\sqrt5}.$

Using $|k|=\sqrt{30}$, $d=\frac{\sqrt{30}}{\sqrt5}=\sqrt6.$

Final answer

$\boxed{\sqrt6}$

Hence the correct option is B.