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 ?
Best Answer
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
$$|RX||XY|=(1+|O'X|)(1-|O'X|)=1-|O'X|^2=1-\frac{(\sqrt2-1)^2}2=\sqrt2-\frac12\implies$$
$$|AB|^2=4\sqrt2-2\implies |AB|=\sqrt{4\sqrt2-2}$$