Explanation

(c)
Given equation is
$\lambda x^{2} + (2\lambda - 3)y^{2} - 4x - 1 = 0$
Here, $a = \lambda,\ b = (2\lambda - 3)$
It represents a circle, if $a = b$
$\Rightarrow \ \lambda = 2\lambda - 3$
$\Rightarrow \ \lambda = 3$
Also, $h = 0$
Then, equation becomes
$3x^{2} + 3y^{2} - 4x - 1 = 0$
$\Rightarrow \ \ x^{2} + y^{2} - \frac{4}{3}x - \frac{1}{3} = 0$
Here, $g = - \frac{2}{3},\ c = - \frac{1}{3},\ f = 0$
$\therefore$ Radius$= \sqrt{\left( - \frac{2}{3} \right)^{2} + 0 - \left( - \frac{1}{3} \right)} = \sqrt{\frac{4}{9} + \frac{1}{3}}$
$= \frac{\sqrt{7}}{3}$