Are you in or are you out?

The equation of the circle is $(x + 1)^{2} + (y + 1)^{2} = r^{2}$ . It will then intercept the positive $y$ -axis when

$1 + (y + 1)^{2} = r^{2} \Longrightarrow y = \sqrt{r^{2} - 1} - 1$ .

(At this point I should mention that this intercept value must be less than $2$ , since if it were $2$ or greater then, because of the curvature of the circle compared to that of the square's diagonal, the area of the portion of the square within the circle would exceed half the total area of the square.)

By symmetry the circle will intercept the positive $x$ -axis at this value, call it $k$ , as well. Now label points $E(0,k), F(k,0)$ , and let $\angle EOA = \theta$ .

Now as $OA = \sqrt{2}, OE = r$ and $AE = k$ , we can use the Cosine Law on triangle $\Delta EOA$ to find that

$k^{2} = r^{2} + 2 - 2\sqrt{2}*r\cos(\theta) \Longrightarrow \cos(\theta) = \dfrac{r^{2} + 2 - k^{2}}{2\sqrt{2}*r} = \dfrac{1 + \sqrt{r^{2} - 1}}{\sqrt{2}*r}$ ,

where I have substituted for $k$ and simplified.

Now the area of $\Delta EOA$ will be $\frac{k}{2}$ , so by symmetry the combined areas of $\Delta EOA$ and $\Delta FOA$ will be $k$ . Also, the area of (circular) sector $AEF$ is $\frac{1}{2} r^{2} * 2\theta = r^{2} * \theta$ . Thus we are looking for $r$ such that

area(sector $AEF$ - (area( $\Delta EOA$ ) + area( $\Delta FOA$ )) = $2$

$\Longrightarrow r^{2}\cos^{-1}(\dfrac{1 + \sqrt{r^{2} - 1}}{\sqrt{2}*r}) - (\sqrt{r^{2} - 1} - 1) = 2$ .

Solving this using numerical methods, (I just used WolframAlpha), we get a value of $r = 2.930622259.....$ , making $\lfloor 1000*r \rfloor = \boxed{2930}$ .

I can't see a way of avoiding using numerical methods at the end, but I think that getting a 'reasonable' equation in $r$ is challenge enough. I preferred this method to a calculus approach, but that method would be quite valid as well.