A) \[\pi /2\]
B) \[\pi /3\]
C) \[\pi /4\]
D) \[\pi /6\]
E) \[\pi /12\]
Correct Answer: B
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\[=\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}\]
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