Explanation

(c)
The equation of the family of circles of radius $r$ is
$(x - a)^{2} + (y - b)^{2} = r^{2}$…..(i)
Where $a$ and $b$ are arbitrary constants
$\Rightarrow$ $2(x - a) + 2(y - b)\frac{dy}{dx} = 0$
$\Rightarrow$ $(x - a) + (y - b)\ \frac{dy}{dx} = 0\$ …(ii)
$\Rightarrow$ $1 + (y - b)\frac{d^{2}y}{dx^{2}} + \left( \frac{dy}{dx} \right)^{2} = 0$
$\Rightarrow$ $(y - b) = - \frac{1 + \left( \frac{dy}{dx} \right)^{2}}{\frac{d^{2}y}{dx^{2}}}$ …(iii)
From eq. (ii),
$(x - a) = \frac{\left\lbrack 1 + \left( \frac{dy}{dx} \right)^{2} \right\rbrack\ \frac{dy}{dx}}{\frac{d^{2}y}{dx^{2}}}$ …(iv)
On putting the value of $(y - b)$ and $(x - a)$, in eq. (i), we get
$\frac{\left\lbrack 1 + \left( \frac{dy}{dx} \right)^{2} \right\rbrack\ \left( \frac{dy}{dx} \right)^{2}}{\left( \frac{d^{2}y}{dx^{2}} \right)^{2}} + \frac{\left\lbrack 1 + \left( \frac{dy}{dx} \right)^{2} \right\rbrack^{2}}{\left( \frac{d^{2}y}{dx^{2}} \right)^{2}} = r^{2}$
$\Rightarrow$ $\left\lbrack 1 + \left( \frac{dy}{dx} \right)^{2} \right\rbrack^{3} = r^{2}\left\lbrack \frac{d^{2}y}{dx^{2}} \right\rbrack^{2}$