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  4. Two tangents are drawn from a point P to the circle x2+y2-2 x-4 y+4=0, such that the angle between these tangents is tan -1((12/5)), where tan -1((12/5)) ∈(0, π) . 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 Δ PAB and Δ CAB is :

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Q. 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 $C$ and these tangents touch the circle at points $A$ and $B$, then the ratio of the areas of $\Delta PAB$ and $\Delta CAB$ is :

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Solution:

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