Question

The differential equation representing the family of circles of constant radius $r$ is

• A. $r^{2} y^{''}=\left[1+\left(y^{'}\right)^{2}\right]^{2}$
• B. $r^{2}\left(y^{''}\right)^{2}=\left[1+\left(y^{'}\right)^{2}\right]^{2}$
• C. $r^{2}\left(y^{''}\right)^{2}=\left[1+\left(y^{'}\right)^{2}\right]^{3}$
• D. $\left(y^{''}\right)^{2}=r^{2}\left[1+\left(y^{'}\right)^{2}\right]^{2}$

Answer 1

Answer: (C)

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{d x}{d \theta}=-r \,\sin \,\theta, \frac{d y}{d \theta}=r \,\cos \,\theta $
$\Rightarrow \frac{d y}{d x}=-\cot \theta$
$\frac{d^{2} y}{d x^{2}}=\text{cosec}^{2} \,\theta \frac{d \theta}{d x}=\frac{-\text{cosec}^{3} \,\theta}{r}$
$\frac{d^{2} y}{d x^{2}}=\frac{-\left(1+\cot ^{2} \,\theta\right)^{3 / 2}}{r} $
$r y^{''}=-\left(1+\left(y^{'}\right)^{2}\right)^{3 / 2}$
Squaring on both sides, we get
$ r^{2}\left(y^{''}\right)^{2}=\left(1+\left(y^{'}\right)^{2}\right)^{3}$