Question

Tangents are drawn from the point $(a,a)$ to the circle $x^2+y^2-2x-2y-6=0$. If the angle between the tangents lies in the range $(\frac{\pi }{3},\pi )$, then the exhaustive range of values of $a$ is

• A. $(1,\infty )$
• B. $(-5,-3)\cup (3,5)$
• C. $(-\infty ,2\sqrt{2})\cup (2\sqrt{2},\infty )$
• D. $(-3,-1)\cup (3,5)$

Answer 1

The correct option is D $(-3,-1)\cup (3,5)$

The point $(a,a)$ lies outside the circle $(1,1),2\sqrt{2}$
$\therefore S^{\prime }>0$
or $2a^2-4a-6>0$ or $a^2-2a-3>0$
$(a+1)(a-3)>0$
$\therefore a<-1$ or $a>3$
If $\theta$ be the angle between the tangents than from the figure
$\tan \frac{\theta }{2}=\frac{r}{\sqrt{S^{\prime }}}=\frac{2\sqrt{2}}{\sqrt{2a^2-4a-6}}\left(g\right).....\left(2\right)$
Since $\frac{\pi }{3}<\theta <\pi \Rightarrow \frac{\pi }{6}<\frac{\theta }{2}<\frac{\pi }{2}$
$\tan \frac{\pi }{6}<\tan \frac{\theta }{2}<\tan \frac{\pi }{2}$
$\therefore \tan \frac{\theta }{2}$ lies in $(\frac{1}{\sqrt{3}},\infty )$
$\therefore \frac{2\sqrt{2}}{\sqrt{2a^2-4a-6}}>\frac{1}{\sqrt{3}}$ by $\left(2\right)$
or $a^2-2a-15<0$ or $(a+3)(a-5)<0$
$\therefore -3<a<5.....\left(3\right)$
Hence from $\left(1\right)$ and $\left(3\right)$, we get
$aϵ(-3,-1)\cup (3,5)$.