Form the differential equation of the family of curve represented by $(2x + a)^2 + y^2 = a^2$, $a$ being parameter.

$2xy\frac{dx}{dy}-y^2+4x^2=0$

The correct answer is Option (3) → $2xy\frac{dx}{dy}-y^2+4x^2=0$

Given $(2x + a)^2 + y^2 = a^2$, $a$ is arbitrary constant   ...(1)

Differentiating it w.r.t. x, we get

$2(2x+a). 2+ 2y \frac{dy}{dx}⇒ 2x+a=-\frac{1}{2}y\frac{dy}{dx}$

$⇒a=-2x-\frac{1}{2}y\frac{dy}{dx}$.

Putting this value of a in (1), we get $\left(-\frac{1}{2}y\frac{dy}{dx}\right)^2 + y^2 = \left(-2x -\frac{1}{2}\frac{dy}{dx}\right)^2$

$⇒\frac{1}{4}y^2(\frac{dy}{dx})^2+ y^2 = 4x^2 + 2xy \frac{dy}{dx} + \frac{1}{4} y^2 (\frac{dy}{dx})^2$

$⇒2xy\frac{dy}{dx}-y^2+4x^2 = 0$, which is the required differential equation of the given family of curve.