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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 ? Figure

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Does "inscribed in a circular region" mean, for example, that the quarter circle's radius intersecting the cord $\,AB\,$ intersects the quarter circle at the tangency point with the (inner) circle? –  DonAntonio Aug 14 '13 at 14:13
    
@DonAntonio Yes , it looks like it. –  A Googler Aug 15 '13 at 8:29
    
Well, it is very important to be sure, @A Googler, since the answers seem to be assuming the slanted line through $\;O\;$ passes through the circle's center $\;O'\;$ and is perpendicular to $\;AB\;$, both these things that are far from being true in general and if the above isn't true... –  DonAntonio Aug 15 '13 at 9:07

5 Answers 5

up vote 2 down vote accepted

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

$$|RX||XY|=(1+|O'X|)(1-|O'X|)=1-|O'X|^2=1-\frac{(\sqrt2-1)^2}2=\sqrt2-\frac12\implies$$

$$|AB|^2=4\sqrt2-2\implies |AB|=\sqrt{4\sqrt2-2}$$

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I think this one is the a rather long answer posted here, Don. :+) –  Babak S. Aug 15 '13 at 13:28
    
Indeed so, @BabakS. , yet I think that either one can assume all those rather basic theorems from Euclidean Geometry or else add them to have an almost self-contained answer. –  DonAntonio Aug 15 '13 at 13:35
    
Thanks for the detailed answer! It taught me a lot. Here is how I think OO'=$\sqrt2$ : Drop a perpendicular O'K from O' to PQ . Then angle O and angle O' are both 45 deg so that OK=O'K=1 and by Pythagorean theorem OO'=$\sqrt2$ . But I don't think this is a good method so can you please tell me how is OO'=$\sqrt2$? –  A Googler Aug 15 '13 at 15:50
    
You were close: draw the lines joining $\,O'\;$ with both $\,OP\;,\;OQ\;$ . This lines are perpendicular to $\,OP\,,\,OQ\;$ respectively (why?) and thus you get a little square there of side one, and its diagonal is precisele $\,OO'\,$ ... –  DonAntonio Aug 15 '13 at 17:07

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, $$|\overline{RT}||\overline{ST}|=|\overline{AT}||\overline{BT}| \quad \left(= \frac{1}{4}|\overline{AB}|^2\right)$$

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Why is that equal to 1/4AB^2 ? Also I don't know RT , ST , AT , BT. . –  A Googler Aug 15 '13 at 6:22
    
@AGoogler: (1) $|AT| = |BT| = 1/2 |AB|$. (2) If $U$ is the center of the smaller circle, then $|RT|$ and $|ST|$ are $|OU|\pm 1$ (depending on which endpoint you call $R$ and which $S$). –  Blue Aug 15 '13 at 6:42
    
Why does AT=BT ? (2) Why ? Sorry but I don't understand . Please explain. –  A Googler Aug 15 '13 at 8:14

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 $AB$. ($O'X=OX-OO'=\cdots$)

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I don't understand . Why is that fact true ? What if the segments didn't intersect ? Also I don't know OB. –  A Googler Aug 15 '13 at 6:20
    
(1) By $OO'$ I meant the lines that passes through $O$ and $O'$ (2) $OB$ was a typo, it should've been $O'B$ which is 1. I've corrected the typo. (3) You only need to $\angle O'XB$ is a right angle and $X$ is the midpoint of $AB$. Then you just write the Pythagoras' theorem for $\Delta O'XB$. –  S.B. Aug 15 '13 at 11:19

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.

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Nice method ! Though it was a bit hard to find the intersection and then apply the distance formula. –  A Googler Aug 15 '13 at 16:03

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 $|OX| = \frac{\sqrt{2}+1}{2}$.

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But how did you find OO' and OX ? –  A Googler Aug 15 '13 at 8:01
    
Is O'OQ = 45 deg ? If so how to prove it ? –  A Googler Aug 15 '13 at 8:28

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