The maximum distance

Suppose we know $Q.$ Then the point $P$ on the circle that maximizes $PQ$ is a point such that $\overline{PQ}$ is normal to the circle at $P.$ This means that it goes through the center of the circle, which is $(0,1).$

The same reasoning holds true if we fix $P$ ; the point $Q$ on the ellipse that maximizes $PQ$ is a point such that $\overline{PQ}$ is normal to the ellipse at $Q.$ If $Q=(x,y),$ the slope of the tangent line to the ellipse at $Q$ is $-x/4y$ by implicit differentiation, so the line $\overline{PQ}$ must have slope $4y/x.$

Putting these together, we get $\frac{4y}x = \frac{y-1}{x-0},$ so either $x=0$ or $y = -\frac13.$ If $x=0,$ the maximum distance $PQ$ is $2+\sqrt{3}.$ We will see that we can do better than this in the other case.

If $y=-\frac13$ then $x = \pm \frac{4\sqrt{2}}3$ ; wlog take $x=\frac{4\sqrt{2}}3.$ So $Q = \left(\frac{4\sqrt{2}}3, -\frac13\right).$

Then we know the slope of $\overline{PQ}$ and its $y$ -intercept, so the line $\overline{PQ}$ is given by the equation $y = -\frac1{\sqrt{2}} x + 1.$ This intersects the circle at the points $(\sqrt{2}, 0)$ and $(-\sqrt{2}, 2).$ The latter clearly gives a larger distance than the former, so $P = (-\sqrt{2},2).$ Computing the distance $PQ$ gives $A = \frac7{\sqrt{3}},$ so that $\lfloor 10000A \rfloor = \fbox{40414}.$