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 $\frac{\pi }{2}<\theta <\pi$, lies in

• A. $(-1,0)\cup (\sqrt{2},5)$
• B. $(-1,0)\cup (1,\sqrt{2})$
• C. $(-\sqrt{2},-1)\cup (0,\sqrt{2})$
• D. $(-\sqrt{2},-1)\cup (1,\sqrt{2})$

Answer 1

The correct option is D $(-\sqrt{2},-1)\cup (1,\sqrt{2})$

Given equation of circle is $x^2+y^2=1$
$C\equiv (0,0)$ and $r=1$
Angle between pair of tangents is given by
$\theta =2{\tan }^{-1}\left(\frac{r}{\sqrt{S_1}}\right)$
$\sqrt{S_1}=\sqrt{a^2+0^2-1}$
$\Rightarrow \sqrt{S_1}=\sqrt{a^2-1}\cdots \left(1\right)$
For $\left(1\right)$ to be defined,
$a^2-1\ge 0$
$\Rightarrow (a+1)(a-1)\ge 0$
$\Rightarrow a\in (-\infty ,-1]\cup [1,\infty )\cdots \left(2\right)$

Now, $\theta =2{\tan }^{-1}\frac{1}{\sqrt{a^2-1}}$
Since $\frac{\pi }{2}<\theta <\pi ,$
$\Rightarrow \frac{\pi }{2}<2{\tan }^{-1}\frac{1}{\sqrt{a^2-1}}<\frac{\pi }{1}$
$\Rightarrow \frac{\pi }{4}<{\tan }^{-1}\frac{1}{\sqrt{a^2-1}}<\frac{\pi }{2}$
$\Rightarrow 1<\frac{1}{\sqrt{a^2-1}}<\infty$
$\Rightarrow 0<\sqrt{a^2-1}<1$
$\Rightarrow 0<a^2-1<1$
$\Rightarrow a^2-2<0$ and $a^2-1>0$
$\Rightarrow (a+\sqrt{2})(a-\sqrt{2})<0$ and $(a+1)(a-1)>0$
$\Rightarrow a\in (-\sqrt{2},\sqrt{2})$ and $a\in (-\infty ,-1)\cup (1,\infty )$
$\Rightarrow a\in (-\sqrt{2},-1)\cup (1,\sqrt{2})\cdots \left(3\right)$
On taking intersection of $\left(2\right)$ and $\left(3\right)$, we get
$a\in (-\sqrt{2},-1)\cup (1,\sqrt{2})$