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 TP.

Answer 1

Join OP and OT

Let OT intersect PQ at a point R.

Then, $TP=TQand\angle PTR=\angle QTR$.

$\therefore TR\perp PQ$

and TR bisects PQ.

$\therefore PR=RQ=4cm$.

Also $OR=\sqrt{{OP}^2-{PR}^2}=\sqrt{5^2-4^2}$ cm

$=\sqrt{25-16}cm=\sqrt{9}cm=3cm$.

Let $TP=xcm$

and $TR=ycm$

From right $\Delta TRP$, we get

${TP}^2={TR}^2+{PR}^2$

$\Rightarrow x^2=y^2+16$

$\Rightarrow x^2-y^2=16\dots \dots \left(i\right)$

From right $\Delta OPT$, we get

${TP}^2+{OP}^2={OT}^2$

$\Rightarrow x^2+5^2={(y+3)}^2[{OT}^2={(OR+RT)}^2]$

$\Rightarrow x^2-y^2=6y-16\dots \dots \left(ii\right)$

From (i) and (ii) , we get

$6y-16=16\Rightarrow 6y=32\Rightarrow y=\frac{16}{3}$

Putting $y=\frac{16}{3}$ in (i) we get

$x^2=16+{\left(\frac{16}{3}\right)}^2=(\frac{256}{9}+16)=\frac{400}{9}$

$\Rightarrow x=\sqrt{\frac{400}{9}}=\frac{20}{3}$

Hence , lenght $TP=xcm=6.67cm$