The minimum number of circles in order to obtain a COVER of a specific square Suppose a unit square $X$, with side length $l=1$ as below, which is COVERed by a set $Y$ of circles with the same constant radius of $r=\dfrac{\sqrt{2}}{10}$, where a cover set $Y$ of another set $X$ means, $x\in Y,\forall x\in X$. 
Questions are :
1) how to obtain the minimum number $n^*$ of circles in $Y$?
2) And I am also interested in the relationship between the ratio $\eta=\dfrac{r}{l}$ of circle radius to the side length of the square, and the number of circles $n^*$.

I also guess, if the pattern as above is right, then $n^*=\left\lceil {\dfrac{\sqrt{2}}{2\eta}} \right\rceil^2 $, but don't know how to prove or disprove it. Probably, below is a counter example?

 A: According to Table 1 of following paper by GK Das et al (2006),
$\hspace0.3in$Efficient algorithm for placing a given number of base stations to cover a convex region
the minimum radius for 23 circles to over a unit square is at most $0.14124482238793135951$. Since this is smaller than $\displaystyle\;\frac{\sqrt{2}}{10}$, at most 23 circles is enough.
I have no idea how the configuration for 23 circles look like. However, following is a
page
which has the configuration of best known covering of a square by up to $12$ equal circles.
As your can see, the configuration doesn't seem to follow any obvious pattern. It is highly
unlikely that $n^{*}$ has any simple formula.
Update
The configurations in Table 1 are computed in a research notes (Ref 15 in GK Das' paper)
$\hspace0.3in$Covering a square with up to 30 equal circles by K.J. Nurmela, P.R.J. Ostergard
An online copy (in postscript) is available here. It has pictures for configuration up to 30 circles.
Based on above paper, following is one way to cover the unit square with 23 circles
of radius $\frac{\sqrt{2}}{10}$.
For simplicity of presentation, we will center the unit square at the origin and
only show the $8$ circles on the first quadrant.

In this configuration, the centers of the circles are positioned at:
$$\begin{array}{|c:l|}
\hline
\text{eg.} & (x,y)\\
\hline
A, B, C & (0,0), (\pm 0.25,0), (\pm 0.5,0)\\
D, E & (\pm 0.125,\pm 0.20756513901392), (\pm 0.375,\pm0.20756513901392)\\
F & (0, \pm 0.41513027802785)\\
G & (\pm 0.22172066050239,\pm 0.40940831933773)\\
H & (\pm 0.41515845958082,\pm 0.38685446089527)\\
\hline
\end{array}$$
The centers of the middle 3 layers (i.e those in the same layers as $A,B,C,D,E$) are forming a triangular lattice elongated in the vertical direction.
A: Well just few thoughts for the problem. The formula for dimension $l=ne^d$ the area that covers a circle is $ E=πρ^2$ and for Square is $E=a^2$ so if we try solve this in order to find the radius, then we take the following result this is if you want a constant radius otherwise if you want multiple circles you have to find the sum $ \sum_{i=1}^n ρ_i =\pm a /\sqrt{\pi} $
for interpretation you can use formulas of Gregory and Leibniz or Abraham Sharp
a = value of square side 
ρ = radius 
π = 3,14...
Ε = Area
l = length 
n = number of shapes 
d = dimension
e = the characteristic value of the shape (radius for circle , side for square)
