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. $2:1$
• D. $3:1$

Answer 1

The correct option is B $9:4$

Given : $x^2+y^2-2x-4y+4=0$
$r=\sqrt{1+4-4}=1\tan \theta =\frac{12}{5}$


Respective areas are
$\Delta PAB=\frac{1}{2}\times L^2\sin \theta \Delta CAB=\frac{1}{2}\times r^2\sin (\pi -\theta )\Rightarrow \frac{\Delta PAB}{\Delta CAB}={\left(\frac{L}{r}\right)}^2\Rightarrow \frac{\Delta PAB}{\Delta CAB}={\cot }^2\frac{\theta }{2}\Rightarrow \frac{\Delta PAB}{\Delta CAB}=\frac{2{\cos }^2\frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}\Rightarrow \frac{\Delta PAB}{\Delta CAB}=\frac{1+\cos \theta }{1-\cos \theta }\Rightarrow \frac{\Delta PAB}{\Delta CAB}=\frac{1+\frac{5}{13}}{1-\frac{5}{13}}=\frac{9}{4}$