JEE PYQ: Circle Question 6
Question 6 - 2021 (17 Mar Shift 2)
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 $\triangle PAB$ and $\triangle CAB$ is:
(1) $11 : 4$
(2) $9 : 4$
(3) $3 : 1$
(4) $2 : 1$
Show Answer
Answer: (2)
Solution
Circle center $C(1,2)$, $r = 1$. $\tan\theta = \frac{12}{5}$, so $PA = \cot\frac{\theta}{2}$. Using $\sin\theta = \frac{12}{13}$, $\cos\theta = \frac{5}{13}$: $\frac{\text{Area of } \triangle PAB}{\text{Area of } \triangle CAB} = \frac{PA}{CA} \cdot \frac{PA}{1} / \frac{CA}{1} = \frac{9}{4}$.
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