Question

Two concentric circles are of radii $5$ cm and $3$ cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer 1

Let two concentric circles be $C_1$ and $C_2$ with centre $O$

$AB$ be the chord of the larger circle $C_2$ which touches the smaller circle $C_1$ at point $P$

Connecting $OP,OA$ and $OB$

$OP=$Radius of smaller circle$=3$cm

$OA=OB=$Radius of larger circle$=5$cm

Since $AB$ is tangent to circle $C_1$

$OP$ is perpendicular to $AB$ since tangent at any point of the circle is perpendicular to the radius through the point of contact.

$\therefore \angle OPA=\angle OPB={90}^{\circ }$

Using Pythagoras theorem

In $△OAP,{OA}^2={OP}^2+{AP}^2\Rightarrow {AP}^2={OA}^2-{OP}^2=25-9=16$

$\therefore AP=4$cm

In $△OPB,{OB}^2={OP}^2+{PB}^2\Rightarrow {PB}^2={OB}^2-{OP}^2=25-9=16$

$\therefore BP=4$cm

Hence $AB=AP+PB=4+4=8$cm