Explanation

(b)
We have, equation of circle is
$x^{2} + y^{2} - 8x + 4y + 4 = 0$
On comparing with standard equation of circle
$x^{2} + y^{2} + 2gx + 2fy + c = 0$, we get
$g = - 4,\ f = 2$ and $c = 4$
$\therefore\ \ Coordinates\ of\ the\ centre = ( - g,\ - f)$
$= (4,\ - 2)$
$\therefore Radius\ of\ the\ circle = \sqrt{g^{2} + f^{2} - c}$
$= \sqrt{(4)^{2} + ( - 2)^{2} - 4}$
$= \sqrt{16 + 4 - 4} = 4$
Here, radius of circle is equal to $x$-coordinate of the centre
$\therefore$ Circle touches $y$-axis