Question

In the given figure, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length of TP.

• A. 6.67 cm
• B. 8.2 cm
• C. 9.3 cm
• D. 7.8 cm

Answer 1

The correct option is A 6.67 cm

Mark point $M$, on the intersection of line $OT$ and $PQ$

By Theorem- Centre of the circle lies on the bisector of the angle between two tangents.
As, $OT$ is the bisector so it also bisects $PQ$
Hence, $PM$ = $MQ$ = 4 cm
And $OP$ = 5 cm (given)

Now consider $△$ $OPM$

By Pythagoras theorem,
${OP}^2$ = ${OM}^2+{PM}^2$
$\Rightarrow$ $5^2$ = ${OM}^2$ + $4^2$
$\Rightarrow$ ${OM}^2$ = $5^2$ - $4^2$ = 25 - 16
$\Rightarrow$ $OM$ = $\sqrt{9}$ = 3 cm
Then, tan $\angle$$OPM$ = $\frac{OM}{PM}$
$\Rightarrow$ tan $\angle$$OPM$ = $\frac{3}{4}$

By theorem- If two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$, then $\angle PTQ=2\angle OPQ$

$\angle$$PTQ$ is bisected by the line $OT$
[ $\because$ By Theorem- Centre of the circle lies on the bisector of the angle between two tangents.]
$\therefore$ $\angle$$PTO$ = $\angle$$OPQ$

tan $\angle$$PTO$ = tan $\angle$$OPQ$ = $\frac{3}{4}$

Consider $△$ $PTO$,

tan $\angle$$PTO$ = $\frac{OP}{TP}$

$\frac{3}{4}$ = $\frac{5}{TP}$ (Given as $OP$ = 5 cm)

$\therefore$ $TP$ = 6.67 cm