Finding the length of chord of a circle. I'm trying to solve the following problem:
The ﬁgure below shows that a circle of radius $r = 1$ is inscribed in quarter circular region $OPQ$. Find the length of the chord $AB$.
I thought about it a lot but still couldn't do anything. Is there some theorem I'm missing ?
Edit : There are some answers to this question but I don't really understand them . Can anyone please post a easy to understand ( possibly with a figure ) answer ?

 A: 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
$$|RX||XY|=(1+|O'X|)(1-|O'X|)=1-|O'X|^2=1-\frac{(\sqrt2-1)^2}2=\sqrt2-\frac12\implies$$
$$|AB|^2=4\sqrt2-2\implies |AB|=\sqrt{4\sqrt2-2}$$
A: Sketch: Let $O'$ be the center of the inscribed circle and $X$ be the intersection of $AB$ and $OO'$. Use the fact that $AB^2 = 4(O'B^2-O'X^2)$ to calculate $AB$. ($O'X=OX-OO'=\cdots$)
A: Hint. Let $R$ and $S$ be the endpoints of the smaller circle's diameter shown in the picture, and let $T$ be the point where $\overline{RS}$ meets $\overline{AB}$. By the Power of a Point Theorem, 
$$|\overline{RT}||\overline{ST}|=|\overline{AT}||\overline{BT}| \quad \left(= \frac{1}{4}|\overline{AB}|^2\right)$$
A: It is quite easy to understand using analytic geometry, i.e. find the equations for each line and circle.
Small circle is: $$(x-1)^2+(y-1)^2=1$$
Line through the origin: $$y=x$$
The intersection of the two above is at: $$2(x-1)^2=1$$ $$x_0=y_0=1+1/\sqrt2$$
Thus, the radius of the large circle is: $$\sqrt(x_0^2+y_0^2)$$
Therefore, the equation for the line $PQ$ is: $$y=\sqrt(x_0^2+y_0^2)-x$$
Find the intersection between this line and the small circle, which will give two points, the distance between which is the desired answer.
I hope this is relatively easy to follow.
A: As in S.B.'s answer, let $O'$ be the center of the inscribed circle and $X$ be the intersection of midpoint of $AB$. By the Pythagorean theorem, you know that $|AX|^2 + |O'X|^2 = |O'A|^2 = 1$ since $O'A$ is a radius of the inscribed circle.
Now, it suffices to calculate $|O'X|$, which can be done by showing that $|OO'| = \sqrt{2}$ and $|OX| = \frac{\sqrt{2}+1}{2}$.
