Question

Form the differential equation of all circles which pass through origin and whose centre 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 the centre of the circle with k as its centre.

So, the equation of circle is

$(x-0{)}^2+(y-k{)}^2=k^2$

$x^2+(y-k{)}^2=k^2$

$x^2+y^2-2ky=0$

$\frac{x^2+y^2}{2y}=k$ .....(1)

Differentiating equation 1 w.r.t. x, we get,

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

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

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

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

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

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

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