Question

The differential equation of the system of all circles of radius $r$ in the $xy$ plane is

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

Answer 1

Answer: (C)

The equation of the family of circles of radius r is
$(x-a{)}^2+(y-b{)}^2=r^2$ ...(1)
Where a & b are arbitrary constants.
Since equation (1) contains two arbitrary constants, we differentiate it two times w.r.t x & the differential equation will be of second order.
Differentiating (1) w.r.t. x, we get
$2\left(x-a\right)+2\left(y-b\right)\frac{dy}{dx}=0$
$\Rightarrow \left(x-a\right)+2\left(y-b\right)\frac{dy}{dx}=0...\left(2\right)$
Differentiating (2) w.r.t. x, we get
$1+\left(y-b\right)\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^{{}^2}=0...\left(3\right)$
$\Rightarrow \left(y-b\right)=-\frac{1+{\left(\frac{dy}{dx}\right)}^{{}^2}}{\frac{d^{2}y}{d{x}^2}}...\left(4\right)$
On putting the value of (y - b) in equation(2), we get
$x-a=\frac{\left[1+{\left(\frac{dy}{dx}\right)}^{{}^2}\right]\frac{dy}{dx}}{\frac{d^{2}y}{d{x}^2}}...\left(5\right)$
Substituting the values of (x - a) &(x - b)in (1), we get
$\frac{{\left[1+{\left(\frac{dy}{dx}\right)}^{{}^2}\right]}^{{}^{{}^2}}{\left(\frac{dy}{dx}\right)}^{{}^2}}{{\left(\frac{d^{2}y}{d{x}^2}\right)}^{{}^2}}+\frac{{\left[1+{\left(\frac{dy}{dx}\right)}^{{}^2}\right]}^{{}^{{}^2}}}{{\left(\frac{d^{2}y}{d{x}^2}\right)}^{{}^2}}=r^2$
$\Rightarrow {\left[1+{\left(\frac{dy}{dx}\right)}^{{}^2}\right]}^{{}^{{}^3}}=r^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^{{}^2}$