Question

From a point T, $13cm$ away from the centre O of a circle of radius $5cm$, tangents PT and QT are drawn. What is the length of AB?

• A. $\frac{19}{3}cm$
• B. $\frac{20}{3}cm$
• C. $\frac{40}{13}cm$
• D. $\frac{22}{3}cm$

Answer 1

The correct option is B $\frac{20}{3}cm$

$P{T}^2=O{T}^2-O{P}^2={13}^2-5^2=144$
$\therefore PT=12$
Let $AE=AP=x;TE=13-5=8$
$\therefore A{T}^2=A{E}^2+T{E}^2$
$\Rightarrow (12-x{)}^2=x^2+8^2$ which gives $x=\frac{10}{3}$
$\Rightarrow AB=2.AE=\frac{20}{3}$