Explanation

(a)
Given, $x^{2} + y^{2} = 6x$ …(i)
and $x^{2} + y^{2} + 6x + 2y + 1 = 0$ …(ii)
From Eq. (i), $x^{2} - 6x + y^{2} = 0$
$\Rightarrow \ (x - 3)^{2} + y^{2} = 3^{2}$
$\therefore$ Centre $(3,\ 0),\ r = 3$
From Eq. (ii),
$x^{2} + 6x + y^{2} + 2y + 1 + 3^{2} = 3^{2}$
$\Rightarrow \ \ \ (x + 3)^{2} + (y + 1)^{2} = 3^{2}$
$\therefore$ Centre $( - 3,\ - 1)$, radius=3
Now, distance between centres
$= \sqrt{(3 + 3)^{2} + 1}$
$= \sqrt{37} > r_{1} + r_{2} = 6$
$\therefore$ Circles do not cut each other
$\Rightarrow$ 4 tangents (two direct and two transversal) are possible