Explanation

(a)
Let $S_{1} \equiv x^{2} + y^{2} - 2x - 2y - 7 = 0$
and $S_{2} \equiv x^{2} + y^{2} + 4x + 2y + k = 0$
here, $g_{1} = - 1,\ \ f_{1} = - 1,\ \ c_{1} = - 7,\ \ r_{1} = 3$
$g_{2} = 2,\ \ f_{2} = 1,\ \ \ c_{2} = k$
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Equation of common chord is $S_{1} - S_{2} = 0$
$\Rightarrow \ \ 6x + 4y + 7 + k = 0$ …(i)
$\because\ \ \ 2\left( g_{1}g_{2} + f_{1}f_{2} \right) = c_{1} + c_{2}$
$\therefore\ \ \ 2( - 2 - 1) = - 7 + k\ \ \Rightarrow \ \ k = 1$
$\therefore$ From Eq. (i), $6x + 4y + 8 = 0$
Let $C_{1}M =$ Perpendicular distance from centre$C_{1}(1,\ 1)$ to the common chord $6x + 4y + 8 = 0.$
$\therefore\ C_{1}M = \frac{|6 + 4 + 8|}{\sqrt{6^{2} + 4^{2}}} = \frac{9}{\sqrt{13}}$
Now, $P\mathcal{Q} = 2PM = 2\sqrt{\left( C_{1}P \right)^{2} - \left( C_{1}M \right)^{2}}$
$= 2\sqrt{9 - \left( \frac{9}{\sqrt{13}} \right)^{2}} = \frac{12}{\sqrt{13}}$