JEE PYQ: Circle Question 25
Question 25 - 2019 (08 Apr Shift 2)
The tangent and the normal lines at the point $(\sqrt{3}, 1)$ to the circle $x^2 + y^2 = 4$ and the $x$-axis form a triangle. The area of this triangle (in square units) is:
(1) $\frac{4}{\sqrt{3}}$
(2) $\frac{1}{3}$
(3) $\frac{2}{\sqrt{3}}$
(4) $\frac{1}{\sqrt{3}}$
Show Answer
Answer: (3)
Solution
Tangent at $(\sqrt{3},1)$: $\sqrt{3}x + y = 4$, meets $x$-axis at $(\frac{4}{\sqrt{3}}, 0)$. Normal passes through origin. Area $= \frac{1}{2} \times \frac{4}{\sqrt{3}} \times 1 = \frac{2}{\sqrt{3}}$.
BETA
