Q. The differential equation of all circles passing through the origin and having their centres on the $x$-axis is
A) $x^{2} = y^{2} + xy\ \frac{dy}{dx}$
B) $x^{2} = y^{2} + 3xy\ \frac{dy}{dx}$
C) $y^{2} = x^{2} + 2xy\ \frac{dy}{dx}$
D) $y^{2} = x^{2} - 2xy\ \frac{dy}{dx}$
Explanation
(c)
Let $x^{2} + y^{2} - 2gx = 0$ ….(i)
$\Rightarrow$ $2x + 2y\ \frac{dy}{dx} - 2g = 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ 2g = \left( 2x + 2y\frac{dy}{dx} \right)$
On putting the value of 2g in eq. (i), we get
$x^{2} + y^{2} - \left( 2x + 2y\ \frac{dy}{dx} \right)x = 0$
$\Rightarrow$ $y^{2} = x^{2} + 2xy\ \frac{dy}{dx}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.