Q. Area enclosed by tangent, normal and x-axis at $( \sqrt{3},1)$ to circle $x^2 + y^2 = 4$ is
Solution:
$x^{2}+y^{2} = 4$
tangent $ \sqrt{3}x+y = 4 $
Normal $x-\sqrt{3}y=0$
Area $ = \frac{1}{2} \times\frac{4}{\sqrt{3}} \times1 $
$ = \frac{2}{\sqrt{3}} $
