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}