Explanation

(b)
Given curves are $9x^{2} - 16y^{2} = 144$
and $x^{2} + y^{2} = 9$
Let the equation of common tangent be
$y = mx + c$
Since, $y = mx + c$ is a tangent to $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$
$\therefore c^{2} = 16m^{2} - 9\left( \because c^{2} = a^{2}m^{2} - b^{2} \right)\ \ \ldots(i)$
Similarly, $y = mx + c$, is a tangent to $x^{2} + y^{2} = 9$
$c = 3\sqrt{m^{2} + 1}\ \Rightarrow \ c^{2} = 9\left( 1 + m^{2} \right)\ \ ...(ii)$
From Eqs. (i) and (ii), we get
$16m^{2} - 9 = 9 + 9m^{2}\ \Rightarrow \ \ m^{2} = \frac{18}{7}\ \Rightarrow \ m = 3\ \sqrt{\frac{2}{7}}$
From Eq. (ii), $c^{2} = 9\left( 1 + \frac{18}{7} \right)\ \ \Rightarrow \ \ c^{2} = 9\ \left( \frac{25}{7} \right)$
$\Rightarrow \ c = \frac{\pm 5}{\sqrt{7}}$
Hence, $y = 3\sqrt{\frac{2}{7}}x + \ \frac{15}{\sqrt{7}}$