A circle $$C_1$$ passes through the origin O and has diameter 4 on the positive x-axis. The line $$y = 2x$$ gives a chord OA of circle $$C_1$$. Let $$C_2$$ be the circle with OA as a diameter. If the tangent to $$C_2$$ at the point A meets the x-axis at P and y-axis at Q, then $$QA : AP$$ is equal to

Circle $$C_1$$ passes through the origin $$O(0,0)$$ and has diameter 4 on the positive $$x$$-axis. Its centre is $$(2, 0)$$, radius $$= 2$$, so its equation is $$(x-2)^2 + y^2 = 4$$.

The line $$y = 2x$$ intersects $$C_1$$; substituting gives $$(x-2)^2 + 4x^2 = 4 \implies 5x^2 - 4x = 0 \implies x(5x - 4) = 0,$$ so $$x = 0$$ (point $$O$$) or $$x = \dfrac{4}{5}$$. Hence $$A = \left(\dfrac{4}{5}, \dfrac{8}{5}\right)$$.

The centre of $$C_2$$ is the midpoint of $$OA$$, namely $$\left(\dfrac{2}{5}, \dfrac{4}{5}\right)$$, and its radius is half of $$|OA|$$: $$\dfrac{|OA|}{2} = \dfrac{1}{2}\sqrt{\dfrac{16}{25}+\dfrac{64}{25}} = \dfrac{1}{2} \cdot \dfrac{\sqrt{80}}{5} = \dfrac{2\sqrt{5}}{5}$$.

For a circle with centre $$(h,k)$$, the tangent at $$(x_1, y_1)$$ satisfies $$(x_1 - h)(x - h) + (y_1 - k)(y - k) = r^2.$$ Substituting $$h=\frac{2}{5}$$, $$k=\frac{4}{5}$$, $$(x_1,y_1)=\left(\frac{4}{5},\frac{8}{5}\right)$$ and $$r=\frac{2\sqrt{5}}{5}$$ gives $$\left(\frac{4}{5}-\frac{2}{5}\right)\left(x-\frac{2}{5}\right) + \left(\frac{8}{5}-\frac{4}{5}\right)\left(y-\frac{4}{5}\right) = \frac{4}{5},$$ which simplifies to $$\frac{2}{5}\left(x-\frac{2}{5}\right) + \frac{4}{5}\left(y-\frac{4}{5}\right) = \frac{4}{5}.$$ Multiplying by 5 yields $$2\left(x-\dfrac{2}{5}\right) + 4\left(y-\dfrac{4}{5}\right) = 4,$$ so $$2x - \frac{4}{5} + 4y - \frac{16}{5} = 4 \implies 2x + 4y = 8 \implies x + 2y = 4.$$

Setting $$y=0$$ gives $$x=4$$ so $$P = (4,0)$$, and setting $$x=0$$ gives $$2y=4$$ so $$y=2$$ and $$Q=(0,2)$$.

Then $$QA = \sqrt{\left(\frac{4}{5}\right)^2 + \left(\frac{8}{5}-2\right)^2} = \sqrt{\frac{16}{25}+\frac{4}{25}} = \frac{\sqrt{20}}{5} = \frac{2\sqrt{5}}{5},$$ $$AP = \sqrt{\left(4-\frac{4}{5}\right)^2 + \left(\frac{8}{5}\right)^2} = \sqrt{\frac{256}{25}+\frac{64}{25}} = \frac{\sqrt{320}}{5} = \frac{8\sqrt{5}}{5},$$ so $$QA : AP = \frac{2\sqrt{5}}{5} : \frac{8\sqrt{5}}{5} = 2 : 8 = \boxed{1 : 4}.$$

The answer is Option A.