Explanation

(a)
Centres and radii of the given circles are $C_{1}(0,\ 0),\ r_{1} = 3$
And $C_{2}( - \alpha,\ - 1)$ and $r_{2} = \sqrt{\alpha^{2} + 1 - 1} = |\alpha|$
Since, two circles touch internally,
$\therefore\ \ \ C_{1}C_{2} = r_{1} - r_{2}$
$\Rightarrow \sqrt{\alpha^{2} + 1^{2}} = 3 - |\alpha|$
$\Rightarrow \ \ \ \alpha^{2} + 1 = 9 + \alpha^{2} - 6|\alpha|$
$\Rightarrow \ \ 6|\alpha| = 8$
$\Rightarrow \ \ |\alpha| = \frac{4}{3}$
$\Rightarrow \ \ \ \alpha \pm \frac{4}{3}$