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 $PR$.
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?
Best Answer
Is my solution correct?
Construction : Join $PS$ and $OR$.
$\angle ORQ = 90^o$ (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 $(\mathtt{ii})$ )
$\implies PR^2 = 256 + 105$
$\implies PR^2 = 361$
$\implies PR = \sqrt{361}$
$\implies PR = 19 cm$