Explanation

(a)
Since, the line $y - 3x = 0$ touches the circle
$\therefore$ radius =perpendicular distance from the centre (1, 1) to the tangent
$= \frac{|1 - 3|}{\sqrt{1 + 9}} = \frac{2}{\sqrt{10}}$ …(i)
Let the other equation of tangent which is passing through origin is $y = mx$
Radius $= \frac{|1 - m|}{\sqrt{1 + m^{2}}}$
$\Rightarrow \ \ \frac{4}{10} = \frac{(1 - m)^{2}}{(1 + m^{2})}$
$\Rightarrow \ \ 3m^{2} - 10m + 3 = 0$
$\Rightarrow \ \ (3m - 1)(m - 3) = 0$
$\Rightarrow \ \ m = 3,\frac{1}{3}$
At $m = 3,\ y = 3x$ it is already given
At $m = \frac{1}{3},\ 3y = x$