Question

Two concentric circles are of radii $5cm$ and $3cm$.

Find the length of the chord of the larger circle which touches the smaller circle.

Answer 1

Solve for length of chord:

In $△OAP$ and $△OFP$

$OA=OF$ (radius of same circle)

$OP=OP$ (given)

$\angle APO=\angle FPO=90°$ (angle made by tangent and radius is $90°$)

Therefore, $△OAP\cong △OFP$ (by RHS congruency criterion)

So, $AP=FP$ (by C.P.C.T.)

Now, in $△OAP$

$OA=5cm$ (radius of larger circle)

$OP=3cm$ (radius of smaller circle)

${\left(OA\right)}^2={\left(OP\right)}^2+{\left(AP\right)}^2$ (By Pythagoras theorem)

$$\begin{gathered} \Rightarrow {\left(5\right)}^2={\left(3\right)}^2+{\left(AP\right)}^2 \\ \Rightarrow 25-9={\left(AP\right)}^2 \\ \Rightarrow {\left(AP\right)}^2=16 \\ \Rightarrow AP=4 \end{gathered}$$

Length of chord $=2\times AP=8cm$

Hence, the length of the chord of the larger circle which touches the smaller circle is $8cm$.