Is there anyone interested in solving this problem?

The sum of the diameters of the given circles is $\frac{10}{\pi}>3$. So, if we project the circles on some side of the square, we get get several segments with total length more than $3$, hence some four of them have a common point, say $A$. The line through $A$ and perpendicular to that side of the square (we've projected on) meets the corresponding four of the circles.