Question

In Fig. $8.63$, $AB$ is a chord of length $16cm$ of a circle of radius $10cm$. The tangents at $A$ and $B$ intersect at a point $P$. Find the length of $PA$.

Answer 1

Join $OA$ as follow

And Given $OB=10cm$
$AB=16cm$
$OP$ is perpendicular to $AB$
so, $AB$ is bisected into $2$ equal parts.
So, $AL=LB=\frac{16}{2}=8$
$PA=PB$ [lengths of tangents to a circle from external point are equal.]

In triangle $OLA$
$O{A}^2=L{O}^2+A{L}^2$ (by pythagoras theorem)
$\Rightarrow L{O}^2=O{A}^2-A{L}^2$
$={10}^2-8^2$
$\Rightarrow LO=$ $\sqrt{100-64}$
$=\sqrt{36}$
$\therefore LO=6cm$ and
$tan\angle AOL=\frac{AL}{OL}=\frac{8}{6}=\frac{4}{3}$
Similarly, from $\Delta OAP$

$\Rightarrow \tan \angle AOL=\frac{PA}{OA}$
$\Rightarrow \frac{4}{3}=\frac{PA}{10}$
$\Rightarrow PA=\frac{4}{3}\times 10$
$=\frac{40}{3}$
$\therefore PA=\frac{40}{3}\approx 13.33cm$