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

$x^2+y^2=1$ is a circle with center $(0,0)$ and radius $r=1$

$PA=PB=$ length of tangent from $(a,0)$ to $x^2+y^2=1$
$PA=PB=\sqrt{a^2-1}$
$\therefore$ In triangle $OAP$

$\tan \alpha =\frac{1}{PA}=\frac{1}{\sqrt{a^2-1}}$

$\therefore$ $\tan 2\alpha =\frac{\frac{2}{\sqrt{a^2-1}}}{1-\frac{1}{a^2-1}}=\frac{2\sqrt{a^2-1}}{(a^2-2)}$

Given $\theta =2\alpha$ and $\frac{\pi }{2}<\theta <\pi$

$\Rightarrow \tan 2\alpha <0$

$\frac{2\sqrt{a^2-1}}{(a^2-2)}<0$

So $a$ must belong to $(-\sqrt{2},-1)\cup (1,\sqrt{2})$
Hence, option 'D' is correct.