Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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

KEAMKEAM 2007

Solution:

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+3{{x}^{2}}-4\sqrt{3}x=4 $
$ \Rightarrow $ $ 4{{x}^{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} $