NTSE Stage 2 SAT 2017 Question Paper

Reason :
As TP is tangent and TBA is secant
So ${\mathrm{TP}}^2=\mathrm{TA}\times \mathrm{TB}$
so, $8^2=\mathrm{TA}\times 4\Rightarrow \mathrm{TA}=16\mathrm{units}$
as TA = 16 units so BA = 12 units
Now in △ BTP,∠PBT=90°
so by Pythagoras theorem
${\mathrm{TP}}^2={\mathrm{TB}}^2+{\mathrm{BP}}^2$
⇒ $8^2={4_{}^{}}_{}^2+{\mathrm{BP}}^2$
⇒ $\mathrm{BP}=4\sqrt{3}$ ....(1)
Now in Δ ABP,∠ABP=90°, so by Pythagoras theorem
${\mathrm{AP}}^2={\mathrm{AB}}^2+{\mathrm{BP}}^2$
⇒ ${12}^2+{\left(4\sqrt{3}\right)}^2={\left(2\mathrm{r}\right)}^2\left[\mathrm{As}\mathrm{AP}=2\mathrm{r}\right]$
$\mathrm{r}=4\sqrt{3}\mathrm{units}$
Hence option (1) is correct.