Question

Two concentric circles are of radii $13cm$ and $12cm$. What is the length (in $cm$) of the chord of the larger circle which touches the smaller circle?

Answer 1

Let $C_1,C_2$ be two circles of radius $13,12$ respectively.

Let $r_1=13cm$ and $r_2=12cm$

Draw a chord $AB$ tangent to $C_2$ at point $P$.

Join $O-A$ and $O-B$

$OP=12cm$ ....... (Radius of smaller circle)

$OA=OB=13cm$ ....... (Radius of bigger circle)

$AB$ is tangent to $C_2$ and $OP$ perpendicular AB

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

Using Pythagoras theorem,

$O{A}^2=O{P}^2+A{P}^2$

$⟹A{P}^2=O{A}^2-O{P}^2$

$⟹A{P}^2={13}^2-{12}^2=169-144=25$

$\therefore AP=5cm$

Similarly, $PB=5cm$

$\therefore AB=AP+PB=5+5=10cm$

Hence, the answer is $10$.