It's a pity this example isn't really describe anywhere. Assuming that there's a triple intersection at $(1,2)$, we get the equation
$$k - 3 - \sin\alpha - \cot\alpha\cdot(2 + 2\cos\alpha - k) = 0$$
Presumably you can get more equations by assuming that the tilted squares touch each other and corners of the straight squares, but these seem not to be satisfied by the given $k,\alpha$.
How to find more equations: the tilted squares come in three groups (two, two and one), and all seem to be just touching each other. For each of the groups, one vertex has one known coordinate (the middle group might have a known vertex at $(1,2)$, but I don't know if that's true). You can get equations from the fact that these groups touch certain corners of the straight squares, and from the (apparent) fact that they just touch each other.
The simplest way to get these equations is to define vectors $s,t$ for the sides of the tilted square (they can be expressed in terms of $\alpha$), and now all these touchings amount to pairs of points $p,q$ which are parallel to either $s$ or $t$. So $p-q$ is proportional to $s$ (or $t$), and we get an equation.
Unfortunately, pursuing this route, I get conflicting equations (conflicting in the sense that when I substitute $k$ and $\alpha$, they are contradictory). Maybe you can try and tell us if it works out for you.