Compute $k$, $k$ greater than 1, for which the circles with radius $k$, centered at $(k,k),(k,-k),(-k,k),(-k,-k),$ are exactly tangent to those with radius 1, centered at $(1,1),(1,-1),(-1,1),(-1,-1)$.
Using Desmos, I have an estimate of approximately 5.828, but I don’t know what the closed form is, nor how to determine it.
On
If a circle at $(k,k)$ and $(1,1)$ are tangent the distances between their centres is $k+1$. So you simply need to solve $\sqrt{(k-1)^2 + (k-1)^2} = k+1$, upon which squaring gives $2(k-1)^2 = (k+1)^2$ or $k^2 - 6k + 1=0$, which is now easily solved and gives $k = 3 + 2 \sqrt 2$ (since the negative solution is assumedly not allowed, the circles would internally tangent in this case).
The distance between $(k,k)$ and $(1,1)$ is $k+1$.
$\sqrt{(k-1)^2+(k-1)^2}=k+1$
$\sqrt{2}(k-1)=k+1$
$(\sqrt{2}-1)k=\sqrt{2}+1$
$k=\dfrac{\sqrt2+1}{\sqrt2-1}=(\sqrt2+1)^2=3+2\sqrt2$