Question

The range of values of 'a' such that the angle $\theta$ between the pair of tangents drawn from (a, 0) to the circle $x^2+y^2=1$ satisfies $\frac{\pi }{2}<\theta <\pi$, is

• A. $(1,2)$
• B. $(1,\sqrt{2})$
  
• C. $(-\sqrt{2},-1)$
  
• D. $(-\sqrt{2},-1)\cup (1,\sqrt{2})$

Answer 1

The correct option is D

$(-\sqrt{2},-1)\cup (1,\sqrt{2})$

Equation of pair of tangents by $S{S}_1=T^2$ is $(a^2-1)y^2-x^2+2ax-a^2=0$
If $\theta$ be the angle between the tangents, then
$tan\theta =\frac{2\sqrt{(h^2-ab)}}{a+b}=\frac{2\sqrt{-(a^2-1)(-1)}}{a^2-2}$
$=\frac{2\sqrt{a^2-1}}{a^2-2}$
$\because \theta$ lies in II quadrant, then tan $\theta$ < 0
$\therefore \frac{2\sqrt{a^2-1}}{a^2-2}<0\Rightarrow a^2-1>0and{a}^2-2<0\Rightarrow 1<a^2<2\Rightarrow aϵ(-\sqrt{2},-1)\cup (1,\sqrt{2})$