Question

O is the centre of the circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle at E. If AB is tangent to the circle at E. Find the length of AB, where TP and TQ are tangents to the circle

Answer 1

AE = AP = x (Tangents from same point A)
Also, in $\Delta OPT$,
$O{T}^2=O{P}^2+P{T}^2$
$\Rightarrow {13}^2=5^2+P{T}^2\Rightarrow P{T}^2={13}^2-5^5=18\times 8=9\times 2\times 2^3\Rightarrow PT=\sqrt{9\times 2^4}=3\times 4=12\therefore AT=(12-x)$
In $\Delta AET,x^2+8^2=(12-x{)}^2\Rightarrow \text{/}{x}^2+8^2={12}^2+\text{/}{x}^2-24x\Rightarrow 24x=144-64\Rightarrow 24x=80\Rightarrow x=\frac{80}{24}$
Now, $AB=2AE=2x=2\times \frac{80}{24}$
$=\frac{\text{/}{80}^{20}}{\text{/}{12}^3}=\frac{20}{3}$