Question

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

Answer 1

Let $TR=y$.

Since $OT$ is perpendicular bisector of $PQ$.

$\therefore PR=QR=4cm$

In right triangle $ORP$, we have

$O{P}^2=O{R}^2+P{R}^2$ ------Pythagoras theorem

$⟹O{R}^2=O{P}^2-P{R}^2=52-45=9$

$⟹OR=3cm$

In right triangles $PRT$ and $OPT$, we have

$T{P}^2=T{R}^2+P{R}^2$ ------Pythagoras theorem

and, $O{T}^2=T{P}^2+O{P}^2$

$⟹O{T}^2=(T{R}^2+P{R}^2)+O{P}^2$ [substituting the value of TP^2]

$⟹(y+3{)}^2=y^2+16+25$

$⟹6y=32$

$⟹y=\frac{16}{3}$

$TR=\frac{16}{3}$

$\therefore T{P}^2=(\frac{16}{3}{)}^2+4^2=\frac{256}{9}+16=\frac{400}{9}$

$⟹TP=\frac{20}{3}cm$