Question

$PQ$ is a chord of length $8cm$ of a circle of radius $5cm$. The tangents at $P$ and $Q$ intersect at a point T. Find the length $TR$.

Answer 1

In the given figure,

Join OT

Let OT intersect PQ at R

TP=TQ (tangent from external point)

In $\Delta TPQ$

TP=TQ

So $\Delta TPQ$ is isosceles

Here OT is bisector of $\angle PTQ$

So $OT\perp PQ$ (RT is altitude of isosceles triangle)

So $PR=RQ=\frac{1}{2}PQ$

$=4cm$

In $\Delta ORP$

$(OP{)}^2=(PR{)}^2+(OR{)}^2$ (Pythagoras theorem)

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

hence,

$OR=3cm$

Let $TP=x$

here, $(TP{)}^2=(PR{)}^2+(RT{)}^2$

$x^2=16+(RT{)}^2$.........$\left(1\right)$

since, $TP$ is tangent

$\begin{array}{c}OP\perp TP\\ \therefore \angle OPT={90}^{\circ }\end{array}$

In right triangle $OPT$

$\begin{array}{c}\Rightarrow {\left(OT\right)}^2={\left(OP\right)}^2+{\left(TP\right)}^2\\ \Rightarrow {\left(OR+RT\right)}^2=5^2+x\\ \Rightarrow {\left(3+RT\right)}^2=5^2+x^2\\ \Rightarrow 9+{\left(RT\right)}^2+6\left(RT\right)=25+x^2\\ from\left(1\right)x^2=16+R{T}^2\\ \Rightarrow 9+R{T}^2+6RT=25+16+R{T}^2\\ \Rightarrow 6RT=32\\ hence,\\ \Rightarrow RT=\frac{16}{3}\end{array}$