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 ${\tan }^{-1}\left(\frac{12}{5}\right)\in \left(0,\mathrm{\pi }\right)$ 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 $△PAB$ and $△CAB$ is:

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

Answer 1

The correct option is B

$9:4$

The explanation for the correct answer.

Step 1: Solve for the value of $AP$

Let $\theta ={\tan }^{-1}\left(\frac{12}{5}\right)$

$\Rightarrow \tan \theta =\frac{12}{5}$

$\Rightarrow \frac{2\tan \left(\frac{\theta }{2}\right)}{1–{\tan }^2\left(\frac{\theta }{2}\right)}=\frac{12}{5}$

$\Rightarrow \tan \left(\frac{\theta }{2}\right)=\frac{2}{3}$

$\Rightarrow \sin \left(\frac{\theta }{2}\right)=\frac{2}{\sqrt{13}}$ and $\cos \left(\frac{\theta }{2}\right)=\frac{3}{\sqrt{13}}$

In $\Delta CAP,\tan \left(\frac{\theta }{2}\right)=\frac{1}{AP}$

$\Rightarrow AP=\frac{3}{2}$

Step 2: Solve for the value of $AB$

In $\Delta APM,\sin \left(\frac{\theta }{2}\right)=\frac{AM}{AP},\cos \left(\frac{\theta }{2}\right)=\frac{AM}{AP}$

$$\begin{gathered} \Rightarrow AM=\frac{3}{\sqrt{13}} \\ \Rightarrow PM=\frac{9}{2\sqrt{13}} \\ \therefore AB=\frac{6}{\sqrt{13}} \end{gathered}$$

Step 3: Solve the required area.

$\therefore$ Area of $\Delta PAB=\left(\frac{1}{2}\right)\times AB\times PM$

$\left(\frac{1}{2}\right)\times \left(\frac{6}{\sqrt{13}}\right)\times \left(\frac{9}{2\sqrt{13}}\right)=\frac{27}{26}$

Now,$\phi ={90}^{\circ }–\left(\frac{\theta }{2}\right)$

In $\Delta CAM,$

$\cos \phi =\frac{CM}{CA}$

$\Rightarrow CM=1\times \cos \left[\left(\frac{\mathrm{\pi }}{2}\right)-\left(\frac{\theta }{2}\right)\right]$

$=1\times \sin \left(\frac{\theta }{2}\right)=\frac{2}{\sqrt{13}}$

$\therefore$ Area of $\Delta CAB=\left(\frac{1}{2}\right)\times AB\times CM$

$=\left(\frac{1}{2}\right)\times \left(\frac{6}{\sqrt{13}}\right)\times \left(\frac{2}{\sqrt{13}}\right)=\frac{6}{13}$

$\therefore$[Area of $\Delta PAB$] / [Area of $\Delta CAB$] $=\frac{\left({\displaystyle \frac{27}{26}}\right)}{\left({\displaystyle \frac{6}{13}}\right)}=\frac{9}{4}$

Hence, option(B) is the correct answer.