Question

The differential equation of the family of circles touching $y$-axis at the origin is

• A. $\left(x^{2}+y^{2}\right) \frac{dy}{dx}-2xy = 0$
• B. $(x^{2}-y^{2}) +2xy \frac{dy}{dx}= 0$
• C. $\left(x^{2}-y^{2}\right) \frac{dy}{dx}-2xy = 0$
• D. $\left(x^{2}+y^{2}\right) \frac{dy}{dx}+2xy = 0$

Answer 1

Answer: (B)

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

Answer 2

The correct option is(B): $(x^{2}-y^{2}) +2xy \frac{dy}{dx}= 0$.

Let centre of circle on X-axis be $(h, 0)$
The radius of circle will be h

$\therefore$ The equation of circle having centre $(h, 0)$ and radius$h$is
$(x-h)^{2} + (y-0)^{2} = h^{2}$
$\Rightarrow x^{2}+h^{2}-2hx+y^{2} = h^{2}$
$\Rightarrow x^{2}-2hx+y^{2}=0 \, \dots(i)$
On differentiating both sides w.r.t x, we get
$2x-2h+2y \frac{dy}{dx}=0$
$\Rightarrow h=x+y \frac{dy}{dx}$
On putting $h = x+y \frac{dy}{dx}$ in E (i), we get
$x^{2}-2\left(x+y \frac{dy}{dx}\right) x+y^{2}=0$
$\Rightarrow -x^{2}+y^{2}-2xy \frac{dy}{dx}=0$
$\Rightarrow \left(x^{2}-y^{2}\right)+2xy \frac{dy}{dx}=0$