I'll condition this answer to the following: the slanted line from $\;O\;$ to the quarter-circle (=QC) crosses $\;AB\;$ at $\;X\;$ and intersects QC at $\;Y=$ the tangency point of both circles .
Now some elementary geometry:
** $\;\color{red}{\text{Theorem}}\;$: if two circles are tangent to each other (externally or internally), the line (or its continuation) through both circles' centers passes through the tangency point, and the other way around: if a segment of line passes through one of the circles' centers and through their tangency point then it passes throught the other circle's center
From this theorem we get that $\,OY\,$ indeed passes through the little circle's center, say $\,O'\,$, since QC and circle $\,O'\,$ are tangent to each other
**$\;\color{red}{\text{Theorem}}\;$: Both tangents to a circle from an external point to the circle (obviously) are equal, and the line joining the external point with the circle's center bisects the two tangents' angle.
From the above theorem we get at once that $\;OY\;$ bisects the straight angle: $$ \angle POY=\angle YOQ=45^\circ\;$$
And this is the gist of all this mess: the segments $\,PQ\;,\;OY\;$ can be seen as the diagonals of a square three of which vertices are $\,P\,,\,O\,,\,Q\,$ and thus the intersect each other and are perpendicular to each other, from which it follows at once that $\;\color{green}{OY\perp AB\iff AX=XB}\;$, since
** $\,\color{red}{\text{Theorem}}\;$: in a circle, a radius (or, in fact, any line segment through the circle's center) bisects a cord iff it is perpendicular to it.
Since a square's diagonal's length is always the length of the square's side's length times $\,\sqrt2\;$ , we get at once that
$$\,|OX|=\frac{\sqrt2}2|PO|=\frac{|OY|}{\sqrt2}\;$$
Observe also that $\;|OO'|=\sqrt2\;$ (this is a nice little exercise and I'd like to leave it to you to prove it. As a hint observe carefully the tangency points of the circle $\,O'\,$ with $\,OP\,,\,OQ\,$ ...if after giving it a good thought you don't succeed write back)
Now the final "touch": denoting by $\,R\,$ the first point of intersection of $\;OY\;$ and the circle $\,O'\,$ , we have
$$|OY|=2+|OR|=2+|OO'|-|RO'|=2+\sqrt2-1=\sqrt2+1$$
so that in fact
$$|OX|=\frac{|OY|}{\sqrt2}=\frac{1+\sqrt2}{\sqrt2}\implies |O'X|=|OX|-|OO'|=\frac{\sqrt2-1}{\sqrt2}$$
and using the
** $\;\color{red}{\text{Theorem}}\;$ : If two cords in a circle intersect each other, then the product of the two parts of one cord equals the product of the two parts of the other cord.
we finally get
$$|RX||XY|=|AX||XB|=\frac12|AB|\frac12|AB|=\frac14|AB|^2$$
and
$$|RX||XY|=(1+|O'X|)(1-|O'X|)=1-|O'X|^2=1-\frac{(\sqrt2-1)^2}2=\sqrt2-\frac12\implies$$
$$|AB|^2=4\sqrt2-2\implies |AB|=\sqrt{4\sqrt2-2}$$
In the midst of a competition you can make all sorts of careless mistakes, but you should at least now see your answer cannot be correct, since $16<8\pi$ so your answer is negative, which is obviously undesirable.
When you get a negative area, what happened is that you took away too much; in this case that is true; you took away the white region twice while only adding in the square once.
The error is relatively easy to fix, fortunately. Your construction also took away the shaded region once (one part for each quarter-circle), so if you add back in the entire square then you have
- added shaded region twice
- subtracted shaded region once
- added unshaded region twice
- subtracted unshaded region twice
which means you have added exactly the shaded region, exactly once; that is the desired area. This does agree with the $32-2\pi(16)/4 = 32-8\pi$ of Sry's comment. (As a sanity check, since $\pi<4$ we have that this answer is at least positive).
Best Answer
The coordinates are right, and therefore the distance to the origin is $\sqrt{2(1-r)^2}$, which is $\sqrt{2}(1-r)$, since $r\lt 1$. But this distance is clearly $1+r$. It follows that $$\sqrt{2}(1-r)=1+r.$$ This is a linear equation in $r$. Solve, and perhaps simplify.