$Two \tan gents are drawn from a point P to the circle$ $x^2+y^2-2x-4y+4=0$, $such that the angle between these \tan gents is$ ${\tan }^{-1}\left(\frac{12}{5}\right)$,$where$ ${\tan }^{-1}\left(\frac{12}{5}\right)\in \left(0,\pi \right)$. $If the center of the areas of$  $\Delta PABand\Delta CAB$is

$\begin{array}{l}\text{The given circle is }{x}^2+y^2-2x-4y+4=0\\ \text{Its Center }\left(1,2\right)\text{\hspace{0.17em} and radius is }r=1\text{\hspace{0.17em}unit}\\ \text{Given angle between the tangents drawn from }P{\text{\hspace{0.17em}to the circle is Tan}}^{-1}\left(\frac{12}{5}\right)\\ 2\theta ={\tan }^{-1}\left(\frac{12}{5}\right)\\ \frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{12}{5}\\ \text{cross multiply and then solve the quadratic equation}\\ \tan \theta =\frac{2}{3}\\ \text{In triangle APC, tan}\theta =\frac{r}{AP}\\ AP=\frac{3}{2}\\ \text{Hence the ratios of the areas of triangles PAB and CAB \hspace{0.17em}is }\frac{\frac{1}{2}\cdot PA\cdot PB\cdot \sin P}{\frac{1}{2}\cdot CA\cdot CB\cdot \sin C}\\ \text{Here both angles P and C are complimentary}\\ \text{Hence the ratio is \hspace{0.17em}\hspace{0.17em}}\end{array}$

$\therefore \frac{Ar\left(\Delta PAB\right)}{Ar\left(\Delta CAB\right)}=\frac{PA.PB.\sin P}{AC.BC\sin C}=\frac{\frac{3}{2}.\frac{3}{2}.\frac{12}{13}}{\mathrm{1.1.}\frac{12}{13}}=\frac{9}{4}=9:4$