Question

Form the differential equation of all circles which pass through origin and whose centres lie on Y-axis.

Answer 1

It is given that, circles pass through origin and their centres lie on Y-axis. Let (0,k) be centre of the circle and radius is k.
So, the equation of circle is,

$(x-0{)}^2+(y-k{)}^2=k^2\Rightarrow x^2+(y-k{)}^2=k^2\Rightarrow x^2+y^2-2ky=0\Rightarrow \frac{x^2+y^2}{2y}=k$
On differentiating Eq. (i) w.r.t. x, we get
$\frac{2y(2x+2y\frac{dy}{dx})-(x^2+y^2)\frac{2dy}{dx}}{4y^2}=0\Rightarrow 4y(x+y\frac{dy}{dx})-2(x^2+y^2)\frac{dy}{dx}=0\Rightarrow 4xy+4y^2\frac{dy}{dx}-2(x^2+y^2)\frac{dy}{dx}=0\Rightarrow [4y^2-2(x^2+y^2\left)\right]\frac{dy}{dx}+4xy=0\Rightarrow (4y^2-2x^2-2y^2)\frac{dy}{dx}+4xy=0\Rightarrow (2y^2-2x^2)\frac{dy}{dx}+4xy=0\Rightarrow (y^2-x^2)\frac{dy}{dx}+2xy=0\Rightarrow (x^2-y^2)\frac{dy}{dx}-2xy=0$