Two tangents are drawn from a point $\mathrm{P}$ to the circle $x^{2}+y^{2}-2 x-4 y+4=0$, such that the angle between these tangents is
$\tan ^{-1}\left(\frac{12}{5}\right)$, where $\tan ^{-1}\left(\frac{12}{5}\right) \in(0, \pi)$. If the centre of the circle is denoted by $\mathrm{C}$ and these tangents touch the circle at points $\mathrm{A}$ and $\mathrm{B}$, then the ratio of the areas of $\Delta \mathrm{PAB}$ and $\triangle \mathrm{CAB}$ is :
Correct Option: , 2
$\tan \theta=\frac{12}{5}$
$\mathrm{PA}=\cot \frac{\theta}{2}$
$\therefore$ area of $\Delta \mathrm{PAB}=\frac{1}{2}(\mathrm{PA})^{2} \sin \theta=\frac{1}{2} \cot ^{2} \frac{\theta}{2} \sin \theta$
$=\frac{1}{2}\left(\frac{1+\cos \theta}{1-\cos \theta}\right) \sin \theta$
area of $\Delta \mathrm{CAB}=\frac{1}{2} \sin \theta=\frac{1}{2}\left(\frac{12}{13}\right)=\frac{6}{13}$
$\therefore \frac{\text { area of } \triangle \mathrm{PAB}}{\text { area of } \Delta \mathrm{CAB}}=\frac{9}{4}$