Explanation

(a)
Since, the given line touches the given circle, the length of the perpendicular from the centre (2, 4) of the circle to the line $3x - 4y - k = 0$ is equal to the radius $\sqrt{4 + 16 + 5} = 5$ of the circle
$\therefore\ \frac{3 \times 2 - 4 \times 4 - k}{\sqrt{9 + 16}} = \pm 5$
$\Rightarrow k = 15\ \ \ (\because k > 0)$
Now, equation of the tangent at $(a,\ b)$ to the given circle is
$xa + yb - 2(x + a) - 4(y + b) - 5 = 0$
$\Rightarrow (a - 2)x + (b - 4)y - (2a + 4b + 5) = 0$
If it represents the given line $3x - 4y - k = 0$
Then, $\frac{a - 2}{3} = \frac{b - 4}{- 4} = \frac{2a + 4b + 5}{k} = l$ (say)
$\Rightarrow a = 3l + 2,\ b = 4 - 4l$
and $2a + 4b + 5 = kl$
$\Rightarrow 2(3l + 2) + 4(4 - 4l) + 5 = 15l\ \ \ (\because k = 15)$
$\Rightarrow \ \ l = 1\ \ \Rightarrow a = 5,\ b = 0$
$\therefore\ k + a + b = 15 + 5 + 0 = 20$