Question

The differential equation of all circles which pass through the origin and whose centres lie on $y$-axis is

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

Answer 1

Answer: (A)

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

Equation of a circle is
$x^2+(y-a{)}^2=a^2$ ...(i)
$2x+2y\frac{dy}{dx}-2a\frac{dy}{dx}=0$ ...(ii)
From Eqs. (i) and (ii), we get
$\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$
$\Rightarrow (x^2-y^2)\frac{dy}{dx}-2xy=0$