Question

In fig. 6, AB is a chord of a circle, with centre O, such that AB = 16 cm and radius of circle is 10 cm. Tangents at A and B intersect each other at P. Find the length of PA ? 

Answer 1

It is given that AB is the chord of the circle with centre O.

Now,

OP\[\perp\] AB 

Therefore, OP bisects AB.

AL = BL = 8 cm

Also,

\[OL = \sqrt{\left( OA \right)^2 - \left( AL \right)^2} = \sqrt{{10}^2 - 8^2} = \sqrt{36} = 6 cm\]

Now,

\[\angle\]PAL +\[\angle\]OAL = `90^o`(Tangent is perpendicular to the radius through the point of contact.)

 

\[\angle\]PAL +\[\angle\]APL = `90^o` (Sum of angles in ∆APL is 180

 

\[\angle\]PAL +\[\angle\]OAL = 

Now, in ∆APL and ∆OAL,

\[\angle\] PLA=\[\angle\]OLA `(90^o each)`

\[\angle\] APL =\[\angle\] OLA   `(proved)`

∴ ∆APL ~∆OAL    (AA similarity)

\[\Rightarrow \frac{PA}{OA} = \frac{AL}{OL}\] (Corresponding sides are proportional.)

\[\Rightarrow \frac{PA}{10} = \frac{8}{6}\]
\[ \Rightarrow PA = \frac{40}{3} cm\]
Thus, the length of the tangent PA is

\[\frac{40}{3}\]