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

Question

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

Dr. Abubacker · Accepted Answer

(d)
Equation of pair of tangents is
$\left( a^{2} - 1 ight)y^{2} - x^{2} + 2ax - a^{2} = 0$
If $	heta$ be the angle between the tangents, then
$	an	heta = \frac{2\sqrt{(h^{2} - ab)}}{a + b}$
$= \frac{2\sqrt{- \left( a^{2} - 1 ight)( - 1)}}{a^{2} - 2}$
$= \frac{2\sqrt{a^{2} - 1}}{a^{2} - 1}$
$\because\ 	heta$ lies in II quadrant, the $	an	heta < 0$
$	herefore\ \frac{2\sqrt{a^{2} - 1}}{a^{2} - 2} < 0$
$\Rightarrow \ \ a^{2} - 1 > 0$ and $a^{2} - 2 < 0$
$\Rightarrow \ \ 1 < a^{2} < 2$
$\Rightarrow \ \ a \in \left( - \sqrt{2},\  - 1 ight) \cup (1,\ \sqrt{2})$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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