Suppose there is a 3x3 square grid, each square of side D, and a point A in the middle of the center square.
Find the circle of centre A and radius R which creates equal arcs inside all the external 8 squares.
Note: Circle areas on squared grid is the closest topic I came across.
I am searching for a general expression for R as a function of D.
On
Arc length is proportional to central angle, so all you have to do is divide the central angle into 8 equal segments ($45^\circ$ each) and see if you can construct the appropriate radius.
A continuity argument implies that you can: let's assume that each little square has a side length of 2 (to avoid fractions later). Then if you circumscribe a circle of radius $\sqrt{2}$ around the central square, four of the outside squares will each have a quarter of the circle and the other four will have nothing. The central angles are then $90^\circ$ for four of the squares and $0^\circ$ for the other four.
Now imagine growing the circle so that it becomes as big as it can, so that the sides of the BIG square are tangent to it (the BIG square consists of all 9 of the little squares). As the circle is growing the four central angles that started at $90^\circ$ will decrease and the other four that started at $0^\circ$ will increase. The questions then are: do the former decrease to below $45^\circ$? do the latter increase to above $45^\circ$? and is there a point where they all become equal to $45^\circ$?
When the circle is as big as it can get (still lying inside the BIG square), the radius is 3 in my units so the sine of half the central angle subtended by the portion of the circle in a square that is directly adjacent to the central square will be $1/3$. This angle is a little less than $20^\circ$, so the central angle will be a bit less than $40^\circ$. The other three such angles are equal by symmetry, and so are the four angles subtended by the arcs in the four corner squares: each of the latter must therefore be greater than $45^\circ$, otherwise the total would not add up to a full circle.
So there is a circle such that the arcs in each of the eight squares are all equal: when each of the central angles of the adjacent squares is $45^\circ$, then the central angle of each of the corner squares has to be $45^\circ$ as well, since they have to add up to $180^\circ$ and they are equal by symmetry.
To find the radius of the circle, we need to make the central angle $45^\circ$. In particular, the central angle of the arc in an adjacent square must be $45^\circ$. But the sine of half of the central angle (denoted by $\theta$ below) will be $1/R$. Using the double angle formula for the sine, we get
$$ \sin 2\theta = 2\sin\theta \cos\theta \implies \frac{1}{\sqrt{2}} = 2 \frac{1}{R}\frac{\sqrt{R^2 - 1}}{R}$$
since $\cos\theta = \sqrt{1 - \sin^2\theta}$. This can be turned into a biquadratic equation by squaring:
$$ R^4 - 8 R^2 + 8 = 0$$
whose solutions are
$$ R^2 = \frac{8 \pm \sqrt{64 - 32}}{2} = 4 \pm 2\sqrt{2} \implies R \approx 2.61312592975\ or\ 1.0823922003$$ But we know that R has to be between $\sqrt{2}$ and $3$, so the bigger solution is the one we want: $$ R \approx 2.61312592975$$
Alternatively since $\theta = 22.5^\circ$, we have $$\frac{1}{R} = \sin\theta \approx 0.382683432365 \implies R \approx 1/0.382683432365 \approx 2.61312592975$$
You clearly want $A$ at the center of the grid. The diagram below has $D=3$ because that was convenient in Geogebra. You want $\angle A_1WB_1=\frac \pi 4$. By symmetry the angles like $\angle B_1WC_1$ will also be $\frac \pi 4$ so all eight of the central angles add up to $2\pi$. That means $\angle B_1WD_1=\frac \pi 8$ If we let $y$ be the $y$ coordinate of $B_1$ we can see we want $$\frac{\frac 32}{\frac 92-y}=\tan \frac \pi 8$$ which, according to Alpha, gives $$ y=\frac 32\left(3-\sqrt{\frac {2+\sqrt 2}{2-\sqrt 2}}\right)\approx 0.87868$$
The radius is then $\sqrt{(\frac 92-y)^2+(\frac 32)^2}$