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I've been trying to solve this geometry question for past 2 hours but haven't got the answer yet.

There are two concentric circles or radius $8 cm $ and $13 cm$ with the common center $O$. $PQ$ is the diameter of the larger circle. And $SQ$ is tangent to the inner circle touching it at $R$. We need to find the length of $PR$.

Find the length of chord PR

I have tried it many times. I did find the length of tangent $SQ$ (Using Pythagoras Theorem). But I didn't help with the length of PR. Can anyone please help me out by giving a hint how to proceed?

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Since PQ is a diameter, PSQ is a right angle. You've already got PQ and SQ. Also, SR=RQ. –  Ed Pegg Feb 9 '13 at 5:22
@EdPegg Thanks a lot this solves my problem. –  ShuklaSannidhya Feb 9 '13 at 5:23
@EdPegg Is my answer correct? –  ShuklaSannidhya Feb 9 '13 at 11:01
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3 Answers

You know the length PQ, and you have found the length QR, and you can get the angle PQR since it's the same as OQR and OQR is a right triangle. Then the law of cosines will get you PR.

EDIT: or you could do it the way @Ed says, in the comments. I knew there had to be something simpler!

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Please check my answer. Is it correct? –  ShuklaSannidhya Feb 9 '13 at 11:01
Well, it's the same answer I get. –  Gerry Myerson Feb 9 '13 at 11:13
Thanks for verification. –  ShuklaSannidhya Feb 9 '13 at 13:01
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up vote 1 down vote accepted

Is my solution correct?

Construction : Join $PS$ and $OR$.

enter image description here

$\angle ORQ = 90^o$ (Radius is perpendicular to tangent at the point of contact)

In $\triangle ORQ$,

$ OQ^2 = OR^2 + RQ^2 $ (using Pythagoras Theorem)

$\implies 13^2 = 8^2 + RQ^2$

$\implies RQ = \sqrt{105}$ $...(\mathtt i)$

Now, $SR=RQ$ (perpendicular from the center of the circle to a chord, bisects the chord)

$\implies SR = \sqrt{105}$ $...(\mathtt{ii})$

Adding $(\mathtt i)$ and $(\mathtt{ii})$,

$SR+RQ=2 \sqrt{105}$

$\implies SQ =2 \sqrt{105} $ $...(\mathtt{iii})$

Now, $\angle PSQ= 90^o$ (angle subtended by a diameter)

In $\triangle PSQ$ ,

$PQ^2=PS^2+SQ^2$ (using Pythagoras Theorem)

$\implies 26^2 = PS^2 + (2\sqrt{105})^2$ ($\because SQ= 2\sqrt{105}$, using $(\mathtt{iii})$ )

$\implies 676 = PS^2 + 420$

$\implies PS^2 = 676-420$

$\implies PS = \sqrt{256}$

$\implies PS = 16$ $...(\mathtt{iv})$

Now, In $\triangle PSR$,

$PR^2 = PS^2 + SR^2$ (using Pythagoras Theorem)

$\implies PR^2 = 16^2 + (\sqrt{105})^2$ ($\because SR= \sqrt{105}$, using $(\mathtt{ii})$ )

$\implies PR^2 = 256 + 105$

$\implies PR^2 = 361$

$\implies PR = \sqrt{361}$

$\implies PR = 19 cm$

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yes. Your solution is correct. An alternative method : Take O as the origin. OQ as x-axis. Find the co-ordinates of the points P and Q. Write down the equations of both the circles and an equation of a tangent line to the smaller circle which passes through Q and hence we get the x- coordinate of the point R and substituting this in the equation of the smaller circle we get the y-coordinate of the point R. By the distance formula we have the length of the line segment PR as 19 cm.

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