Explanation

(c)
The given equation of circle is
$x^{2} + y^{2} - 6x + 4y - 12 = 0$
The centre and radius of circle are $(3,\ - 2)$ and 5 respectively
$\therefore$ Length of perpendicular from $(3,\ - 2)$ to $4x + 3y + \lambda = 0$ is equal to radius 5
$\therefore\ \ \left| \frac{12 - 6 + \lambda}{\sqrt{16 + 9}} \right| = 5$
$\Rightarrow 6 + \lambda = \pm 25$
$\Rightarrow \ \lambda = 19,\ - 31$
Then, equations of tangents are $4x + 3y + 19 = 0$ and $4x + 3y - 31 = 0$