Question

The area bounded by the circles $x^2+y^2=1,x^2+y^2=4$ and the pair of lines $\sqrt{3}(x^2+y^2)=4xy,$ is equal to

• A. $\frac{\pi }{2}$
• B. $\frac{5\pi }{2}$
• C. $3\pi$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (D)

$\frac{\pi }{4}$

The angle $\theta$ between the lines represented by

$\sqrt{3}{x}^2-4xy+\sqrt{3}{y}^2=0$ is given by

$\theta ={\tan }^{-1}\frac{\sqrt{h^2-ab}}{|a+b|}$

$={\tan }^{-1}\frac{2\sqrt{2^2-3}}{\sqrt{3}+\sqrt{3}}=\frac{1}{\sqrt{3}}$

Gives $\theta =\frac{\pi }{6}$

Hence, the shaded area $=\frac{\pi /6}{2\pi }\times \pi (2^2-1^2)=\frac{\pi }{4}$