The range of values of a such that the angle | vedclass

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(D) Let the circle be $x^2 + y^2 = 1$. The radius $r = 1$ and the center is $(0, 0)$.Let the point be $P(a, 0)$. The distance from the center to $P$ i

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

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(D) Let the circle be $x^2 + y^2 = 1$. The radius $r = 1$ and the center is $(0, 0)$.Let the point be $P(a, 0)$. The distance from the center to $P$ is $d = |a|$.Let $	heta$ be the angle between the tangents. Then $\sin(	heta/2) = \frac{r}{d} = \frac{1}{|a|}$.Given $\frac{\pi}{2} Taking the sine function,since $\sin(x)$ is increasing in this interval,we get $\sin(\frac{\pi}{4}) This implies $\frac{1}{\sqrt{2}} Taking the reciprocal,we get $1 This means $|a| > 1$ and $|a| Thus,$a \in (-\sqrt{2}, -1) \cup (1, \sqrt{2})$.

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

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