Explanation

(c)
Given, $x^{2} + y^{2} = \frac{1}{5}$
Centre of the circle is (0, 0)
Let equation of tangent which are parallel to $3x + 4y - 1 = 0$ is
$3x + 4y + \lambda = 0$ …(i)
$\therefore\ \ \frac{3 \times 0 + 4 \times 0 + \lambda}{\sqrt{(3)^{2} + (4)^{2}}} = \pm \frac{1}{\sqrt{5}}$
$\Rightarrow \ \lambda = \pm \sqrt{5}$
On putting the value of $\lambda$ in Eq. (i), we get
$3x + 4y = \pm \sqrt{5}$