Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel. 
Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel, and a circle of radius $(2 - \sqrt{2})$ cm covered within these squares. Find the length of $AD$.

What I Tried: Here is a picture :-

I made an attempt solving this and found rather a weird conclusion.
We have that $A,P,C,R$ are collinear, and also $AP = CR$.
Let $DL = LS = BK = KQ = x$ . We have $AP = x\sqrt{2}$ . Since $PC$ = $(4 - 2\sqrt{2})$ , $AS = (4 - 2\sqrt{2} + 2x\sqrt{2})$ .
Now, $PF = FC$ . So by Pythagoras Theorem on $\Delta PFC$ , $PF = FC = \sqrt{12 - 8 \sqrt{2}}$.
$AL = LS = \sqrt{12 - 8 \sqrt{2}} + 2x$ and $AS = (4 - 2\sqrt{2} + 2x\sqrt{2})$ . So now by Pythagoras Theorem :-
$$2\bigg(\sqrt{12 - 8 \sqrt{2}} + 2x\bigg)^2 = (4 - 2\sqrt{2} + 2x\sqrt{2})$$
$$\rightarrow 2\bigg(12 - 8\sqrt{2} + 4\sqrt{12 - 8\sqrt{2}}x + 4x^2\bigg) = 8x^2 + 16\sqrt{2}x - 16x - 16\sqrt{2} + 24$$
$$\rightarrow 24 - 16\sqrt{2} + 8\sqrt{12 - 8\sqrt{2}}x + 8x^2 = 8x^2 + 16\sqrt{2}x - 16x - 16\sqrt{2} + 24$$
$$\rightarrow \sqrt{12 - 8\sqrt{2}}x = 2\sqrt{2}x - 16x$$
$$\rightarrow \sqrt{12 - 8\sqrt{2}} = 2\sqrt{2} - 16$$
But this is obviously false.
So I literally ended up in a completely false equation, might be making a mistake in this complicated calculations, and I cannot find it. So can anyone tell me where I made the mistake, or I am solving this in the correct way or not?

Using the fact that the side of the outer square and diagonal of the inner square are equal to the diameter of the circle,
$AD = \frac{AL + PE}{2} = \frac{2r + r\sqrt2}{2} = \frac{(2 + \sqrt2)(2 - \sqrt2)}{2} = 1$
