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

Answer 1

The Correct Option is C

The equation of the family of circles of radius r is
$(x - a)^2 + (y - b)^2 = r^2$$\quad$$\quad$$\quad$ ...(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
$2\left(x-a\right)+2\left(y-b\right) \frac{dy}{dx}=0$
$\Rightarrow \left(x-a\right)+2\left(y-b\right) \frac{dy}{dx}=0\quad\quad\quad\quad...\left(2\right)$
Differentiating (2) w.r.t. x, we get
$1+\left(y-b\right) \frac{d^{2}y}{dx^{2}}+\left(\frac{dy}{dx}\right)^{^2}=0\quad\quad\quad\quad...\left(3\right)$
$\Rightarrow \left(y-b\right)=-\frac{1+\left(\frac{dy}{dx}\right)^{^2}}{\frac{d^{2}y}{dx^{2}}}\quad\quad\quad\quad\quad...\left(4\right)$
On putting the value of (y - b) in equation(2), we get
$x-a=\frac{\left[1+\left(\frac{dy}{dx}\right)^{^2}\right] \frac{dy}{dx}}{\frac{d^{2}y}{dx^{2}}}\quad\quad\quad\quad\quad...\left(5\right)$
Substituting the values of (x - a) &(x - b)in (1), we get
$\frac{\left[1+\left(\frac{dy}{dx}\right)^{^2}\right]^{^{^2}}\left(\frac{dy}{dx}\right)^{^2}}{\left(\frac{d^{2}y}{dx^{2}}\right)^{^2}}+\frac{\left[1+\left(\frac{dy}{dx}\right)^{^2}\right]^{^{^2}}}{\left(\frac{d^{2}y}{dx^{2}}\right)^{^2}}=r^{2}$
$\Rightarrow \left[1+\left(\frac{dy}{dx}\right)^{^2}\right]^{^{^3}}=r^{2}\left(\frac{d^{2}y}{dx^{2}}\right)^{^2}$