Question

Two tangents are drawn from a point $P$ to the circle $x^2+y^2-2x-4y+4=0$, 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 $C$ and these tangents touch the circle at points $A$ and $B$, then the ratio of the areas of $\Delta PAB$ and $\Delta CAB$ is :

• A. 11 : 4
• B. 9 : 4
• C. 3 : 1
• D. 2 : 1

Answer 1

Answer: (B)

$\tan \theta =\frac{12}{5}$
$PA=\cot \frac{\theta }{2}$
$\therefore$ area of $△PAB=\frac{1}{2}(PA{)}^2\sin \theta =\frac{1}{2}{\cot }^2\frac{\theta }{2}\sin \theta$
$=\frac{1}{2}\left(\frac{1+\cos \theta }{1-\cos \theta }\right)\sin \theta$
$=\frac{1}{2}\left(\frac{1+\frac{5}{13}}{1-\frac{5}{13}}\right)\left(\frac{12}{13}\right)=\frac{118}{218}\times \frac{2}{13}=\frac{27}{26}$
area of $\Delta CAB=\frac{1}{2}\sin \theta =\frac{1}{2}\left(\frac{12}{13}\right)=\frac{6}{13}$
$\therefore \frac{\text{ area of }\Delta PAB}{\text{ area of }\Delta CAB}=\frac{9}{4}$