You are given a square with sidelength $x$. Draw $4$ circles $\Gamma_A$, $\Gamma_B$, $\Gamma_C$, $\Gamma_D$ in the square with radius $1$, such that no two circles touch/intersect. The circle can't intersect the borderline of the square aswell (it can touch though).
Find the minimal value of $x$ such that there doesn't exists points $A_1$, $B_1$, $C_1$ $D_1$ on $\Gamma_A$, $\Gamma_B$, $\Gamma_C$, $\Gamma_D$ respectively such that $A_1B_1C_1D_1$ is a cyclic quadrilateral (in short, find the smallest $x$ such that $A_1B_1C_1D_1$ never is cyclic).
My progress: I conjecture that the answer is $x=6$, and on the following image you find a construction:
How I tackled the problem: I let H,I,J be moving points on $\Gamma_A, \Gamma_B$ and $\Gamma_C$. Then I construct the circle and find all possible points in the plane it could reach. The blue dots/region is the reach of the circle. In this specific construction, it seems that the region never intersects $\Gamma_D$, and therefore I think $x=6$ is the minimum. However, this may not be an optimal construction, so I am in doubt.
edited: A second image to prove that conjecture in comment is false: In the image, we see that ELM is an equilateral triangle. We can show that there doesn't exist another circle in the triangle that can't be reached by three points on circles with circumcenters E,L,M and radius one. The "optimal circle" would've been the one centered at F. However, in the image, we see that there exists a circle intersecting our circle centered at F, proving that this construction isn't optimal.
edited x2: I think I found a better construction, this can lower the bound of x=6. No idea why I haven't tried this construction as it seems like a pretty easy idea: 3 circles, one in each corner. Circle in the middle can't be reached.
