Q. The differential equation of all circles which passes through the origin and whose centre lies on $y$-axis, is
A) $\left( x^{2} - y^{2} \right)\frac{dy}{dx} - 2xy = 0\$
B) $\left( x^{2} - y^{2} \right)\frac{dy}{dx} + 2xy = 0\$
C) $\left( x^{2} - y^{2} \right)\frac{dy}{dx} - xy = 0$
D) $\left( x^{2} - y^{2} \right)\frac{dy}{dx} + xy = 0$
Explanation
(a)
Let $x^{2} + y^{2} - 2ky = 0$
$\Rightarrow$ $2x + 2y\ \frac{dy}{dx} - 2k\ \frac{dy}{dx} = 0$
$\Rightarrow$ $k = \frac{k}{\left( \frac{dy}{dx} \right)} + y$
From Eq. (i),
$x^{2} + y^{2} - 2\left( \frac{x}{(\frac{dy}{dx})} + y \right)y = 0$
$\Rightarrow$ $\left( x^{2} - y^{2} \right)\frac{dy}{dx} - 2xy = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.