Question

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

Answer 1

In right-angled $△OPR$,

${OP}^2={PR}^2+{OR}^2$

$\Rightarrow {OR}^2=5^2-4^2=25-16=9$

$\Rightarrow OR=\sqrt{9}=3$

In right-angled $△PRT$,

${TP}^2={RT}^2+{PR}^2.....\left(1\right)$

In right-angled $△OTP$,

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

${OT}^2={OP}^2+({RT}^2+{PR}^2)\left(\text{From}\left(1\right)\right)$

${(OR+RT)}^2=5^2+{RT}^2+4^2$

${(3+RT)}^2=41+{RT}^2$

${RT}^2+9+6RT=41+{RT}^2$

$\Rightarrow 6RT=41-9$

$\Rightarrow RT=\frac{32}{6}=\frac{16}{3}$

Substituting the value of $RT$ in equation $\left(1\right)$, we have

${TP}^2={\left(\frac{16}{3}\right)}^2+4^2$

$\Rightarrow {TP}^2=\frac{256}{9}+16$

$\Rightarrow TP=\sqrt{\frac{256+144}{9}}$

$\to TP=\sqrt{\frac{400}{9}}=\frac{20}{3}cm$