# Two tangents are drawn

Question:

Two tangents are drawn from a point $P$ 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 A and B, then the ratio of the areas of $\triangle \mathrm{PAB}$ and $\triangle \mathrm{CAB}$ is:

1. 11 : 4

2. 9 : 4

3. 3 : 1

4. 2 : 1

Correct Option: , 2

Solution:

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

$\mathrm{PA}=\cot \frac{\theta}{2}$

$\therefore$ area of $\Delta \mathrm{PAB}=\frac{1}{2}(\mathrm{PA})^{2} \sin \theta=\frac{1}{2} \cot ^{2} \frac{\theta}{2} \sin \theta$

$=\frac{1}{2}\left(\frac{1+\cos \theta}{1-\cos \theta}\right) \sin \theta$

$=\frac{1}{2}\left(\frac{1+\frac{5}{13}}{1-\frac{5}{13}}\right)\left(\frac{12}{13}\right)=\frac{1}{2} \frac{18}{18} \times \frac{2}{13}=\frac{27}{26}$

area of $\Delta \mathrm{CAB}=\frac{1}{2} \sin \theta=\frac{1}{2}\left(\frac{12}{13}\right)=\frac{6}{13}$

$\therefore \frac{\text { area of } \Delta \mathrm{PAB}}{\text { area of } \Delta \mathrm{CAB}}=\frac{9}{4}$