Explanation

(b)
The equation of the circle passing through (2, 0), (0, 1) and (4, 5) is
$3\left( x^{2} + y^{2} \right) - 13x - 17y + 14 = 0$
This passes through $(0,\ c)$
$\therefore\ \ \ 3c^{2} - 17c + 14 = 0\ \ \Rightarrow \ \ c = 1,\frac{14}{3}$
Since, $c = 1$ is already there, for point (0, 1)
Therefore, we take $c = \frac{14}{3}$