Question

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

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

Answer 1

Answer: (C)

Family of circle of constant radius $r$ is
$(x-a{)}^2+(y-b{)}^2=r^2$
Let $x=a+r\cos \theta ,y=b+r\sin \theta$
$\frac{dx}{d\theta }=-r\sin \theta ,\frac{dy}{d\theta }=r\cos \theta$
$\Rightarrow \frac{dy}{dx}=-\cot \theta$
$\frac{d^{2}y}{d{x}^2}={\text{cosec}}^2\theta \frac{d\theta }{dx}=\frac{-{\text{cosec}}^3\theta }{r}$
$\frac{d^{2}y}{d{x}^2}=\frac{-{\left(1+{\cot }^2\theta \right)}^{3/2}}{r}$
$r{y}^{{}^{\prime \prime }}=-{\left(1+{\left(y^{{}^{\prime }}\right)}^2\right)}^{3/2}$
Squaring on both sides, we get
$r^2{\left(y^{{}^{\prime \prime }}\right)}^2={\left(1+{\left(y^{{}^{\prime }}\right)}^2\right)}^3$