Explanation

(b)
The equation of common chord $P\mathcal{Q}$ is
$2x + 1 = 0\ \ \ \ \lbrack ie,\ S_{2} - S_{1} = 0\rbrack$
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Here, $C_{1} = \left( - 1,\frac{- 3}{2} \right),\ \ \ r_{1} = \frac{3}{2} = C_{1}P$
and $C_{2} = \left( - 2,\frac{- 3}{2} \right),\ \ r_{2} = \frac{\sqrt{17}}{2}$
$C_{1}M =$ Perpendicular distance from $C_{1}$ to the common chord
$\therefore\ \ \ C_{1}M = \frac{| - 2 + 1|}{\sqrt{2^{2}}} = \frac{1}{2}$
Now, $P\mathcal{Q} = 2PM = 2\sqrt{\left( C_{1}M \right)^{2} - \left( C_{1}M \right)^{2}} = 2\sqrt{\frac{9}{4} - \frac{1}{4}} = 2\sqrt{2}$