Circles
Circle
star_batch_jee_advanced_2025
Grade None
Question:
The area of region bounded by circle $f(x,y) = 0$ with x-axis in the first quadrant is:
$3 + \frac{25}{8}\left(\pi - \tan^{-1}\frac{1}{2}\right)$
$3 + \frac{25}{8}\left(\tan^{-1}\frac{24}{11}\right)$
$3 + \frac{25}{8}\left(2\pi - \tan^{-1}\frac{3}{4}\right)$
$3 + \frac{25}{8}\left(2\pi - \tan^{-1}\frac{24}{7}\right)$
Step-by-Step Solution
Key Concept: Area calculation requires finding the sector area minus triangular portions using inverse trigonometric relationships.
Given $\theta = \tan^{-1}(\frac{3}{4})$, we find $2\theta = \tan^{-1}(\frac{24}{7})$ using the double angle formula. The area of the region inside circle $f(x,y) = 0$ above the x-axis is computed as $\frac{1}{2}(\frac{5}{2})^2(2\pi - 2\tan^{-1}(\frac{24}{7})) - \frac{1}{2} \times 3 \times 2 - 3 + \frac{25}{8}(2\pi - \tan^{-1}(\frac{24}{7}))$.
Correct Answer: 4