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

The Correct Option is B

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