[Math] Find the length of this chord.

analytic geometrycirclesgeometry

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?

Best Answer

Is my solution correct?

Construction : Join $PS$ and $OR$.

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$\angle ORQ = 90^o$ (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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