Let the lengths of intercepts on $$x$$-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 origin to a tangent to this circle which is perpendicular to the line $$x + 2y = 0$$, is equal to:

The circle is $$x^2 + y^2 + ax + 2ay + c = 0$$ with $$a < 0$$. Comparing with the standard form $$x^2 + y^2 + 2gx + 2fy + c = 0$$, we get $$g = \frac{a}{2}$$ and $$f = a$$. The center is $$\left(-\frac{a}{2}, -a\right)$$ and the radius squared is $$r^2 = g^2 + f^2 - c = \frac{a^2}{4} + a^2 - c = \frac{5a^2}{4} - c$$.

The length of the intercept on the x-axis is given by $$2\sqrt{g^2 - c}$$. Setting this equal to $$2\sqrt{2}$$: $$2\sqrt{\frac{a^2}{4} - c} = 2\sqrt{2}$$, which gives $$\frac{a^2}{4} - c = 2$$ ... (i).

The length of the intercept on the y-axis is given by $$2\sqrt{f^2 - c}$$. Setting this equal to $$2\sqrt{5}$$: $$2\sqrt{a^2 - c} = 2\sqrt{5}$$, which gives $$a^2 - c = 5$$ ... (ii).

Subtracting (i) from (ii): $$a^2 - \frac{a^2}{4} = 3$$, so $$\frac{3a^2}{4} = 3$$, giving $$a^2 = 4$$. Since $$a < 0$$, we have $$a = -2$$.

Substituting back into (i): $$\frac{4}{4} - c = 2$$, so $$c = -1$$.

The center of the circle is $$\left(-\frac{-2}{2}, -(-2)\right) = (1, 2)$$ and $$r^2 = \frac{5 \cdot 4}{4} - (-1) = 5 + 1 = 6$$, so $$r = \sqrt{6}$$.

We need tangent lines perpendicular to $$x + 2y = 0$$, which has slope $$-\frac{1}{2}$$. The perpendicular direction has slope $$2$$. So the tangent lines have the form $$y = 2x + k$$, or equivalently $$2x - y + k = 0$$.

For a tangent to the circle, the distance from the center $$(1, 2)$$ to the line must equal the radius: $$\frac{|2(1) - 2 + k|}{\sqrt{4 + 1}} = \sqrt{6}$$, so $$\frac{|k|}{\sqrt{5}} = \sqrt{6}$$, giving $$|k| = \sqrt{30}$$.

The two tangent lines are $$2x - y + \sqrt{30} = 0$$ and $$2x - y - \sqrt{30} = 0$$. The distance from the origin $$(0, 0)$$ to the line $$2x - y + k = 0$$ is $$\frac{|k|}{\sqrt{5}}$$. For $$k = \sqrt{30}$$, the distance is $$\frac{\sqrt{30}}{\sqrt{5}} = \sqrt{6}$$. For $$k = -\sqrt{30}$$, the distance is also $$\frac{\sqrt{30}}{\sqrt{5}} = \sqrt{6}$$.

Therefore, the shortest distance from the origin to a tangent to this circle perpendicular to $$x + 2y = 0$$ is $$\sqrt{6}$$.