Question

The acute angle between the lines joining the origin to the points of intersection of the line $\sqrt{3}x+y=2$ and the circle is $x^2+y^2=4,$ is

• A. $\pi /2$
• B. $\pi /3$
• C. $\pi /4$
• D. $\pi /6$
• E. $\pi /12$

Answer 1

Answer: (B)

Equation of line is $\sqrt{3}x+y=2$ ...(i) and equation of circle is $x^2+y^2=4$ ...(ii) From Eqs. (i) and (ii), we get $x^2+2{(-\sqrt{3}x)}^2=4$
$\Rightarrow$ $x^2+4+3x^2-4\sqrt{3}x=4$
$\Rightarrow$ $4x^2-4\sqrt{3}x=0$
$\Rightarrow$ $x(x-\sqrt{3})=0$
$\Rightarrow$ $x=0,\sqrt{3}$
$\therefore$ Points of intersection of line and circle are (0,2) and $(\sqrt{3},-1)$ . Slope of line joining (0, 0) and (0, 2)
$=\frac{2-0}{0-0}=\infty \Rightarrow {\theta }_1=\frac{\pi }{2}$ And slope of line joining (0, 0) and $(\sqrt{3},-1)$
$=\frac{-1}{\sqrt{3}}\Rightarrow {\theta }_2=\frac{\pi }{6}$ .
Required angle
$=\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}$