question_answer

The range of values of 'a' such that the angle \[\alpha \] between the tangents drawn from (a, 0) to the circle \[{{x}^{2}}+{{y}^{2}}=4\] which lies in \[\left( \frac{\pi }{3},\,\frac{\pi }{2} \right)\] are (where a is the angle in which circle lies)

A)  \[\left( -4,\,\sqrt{2} \right)\cup \left( \sqrt{2},4 \right)\]

B)  \[\left( -4,\,-2\sqrt{2} \right)\,\cup \,\left( 2\sqrt{2},\,4 \right)\]

C)  \[\left( -4,\,\sqrt{2} \right)\cup \left( 2\sqrt{2},\,4 \right)\]

D)  \[\left( -4,\,4 \right)\]

Correct Answer: B

Solution :

\[\sin \left( \frac{\alpha }{2} \right)=\frac{2}{a}\] \[\alpha \in \left( \frac{\pi }{3},\,\frac{\pi }{2} \right)\] \[\Rightarrow \,\,\frac{\alpha }{2}\in \left( \frac{\pi }{6},\,\frac{\pi }{4} \right)\] \[\Rightarrow \,\,\frac{2}{a}\in \,\left( \frac{1}{2},\,\frac{1}{\sqrt{2}} \right)\] \[\Rightarrow \,\,a\in \,\left( \frac{1}{2},\,\frac{1}{\sqrt{2}} \right)\] If (a, 0) is on left side of origin then \[-\frac{2}{a}\in \left( \frac{1}{2},\,\frac{1}{\sqrt{2}} \right)\] \[\Rightarrow \,\,a\in \left( -4,\,-2\sqrt{2} \right)\] Hence, \[a\in \,\left( -4,\,-2\sqrt{2} \right)\cup \,\left( 2\sqrt{2},\,4 \right)\]