Explanation

(a)
Given equation of circle is
$x^{2} + y^{2} - 2x - 6y + 6 = 0$ …(i)
Its centre is (1, 3) and radius $= \sqrt{1 + 9 - 6} = 2$
Equation of any line through (0, 1) is
$y - 1 = m(x - 0)$
$\Rightarrow \ mx - y + 1 = 0$ …(ii)
If it touches the circle (i), then the length of perpendicular from centre (1, 3) to the circle is equal to radius 2
$\therefore\ \ \frac{m - 3 + 1}{\sqrt{m^{2} + 1}} = \pm 2$
$\Rightarrow \ \ \ (m - 2)^{2} = 4(m^{2} + 1)$
$\therefore\ \ \ \ m = 0,\ - \frac{4}{3}$
On substituting these values of $m$ in Eq. (ii), the required tangent are $y - 1 = 0$ and $4x + 3y - 3 = 0$