Answer:
Option C
Explanation:
Family of circle of constant radius r is
$(x-a)^{2}+(y-b)^{2}=r^{2}$
Let $x=a + r \cos \theta$,y=$b+r \sin \theta$
$\frac{dx}{d\theta}=-r \sin \theta, \frac{dy}{d \theta}= r \cos \theta \Rightarrow \frac{dy}{dx}=-\cot \theta$
$\frac{dy ^{2}}{dx^{2}}=cosec ^{2} \theta, \frac{d\theta}{d x}= \frac{- cosec ^{2} \theta}{r}$
$\frac{dy ^{2}}{dx^{2}}= \frac{-(1+ \cot ^{2} \theta)^{3/2}}{r}$
$ry"= -(1+(y')^{2})^{3/2}$
squaring on both sides , we get
$r^{2} (y")^{2}=[1+(y')^{2}]^{3}$
