Question

Form the differential equation of all circles which touch the x-axis at the origin.

Answer 1

Let r be the readius of all circles which touch the x-axis at origin. So, centre of all such circles must lie on y-axis. Therefore, the centre will be of the form (0,r).
So, the equation of all such circles: $(x-0{)}^2+(y-r{)}^2=r^2$ i.e., $x^2+y^2-2ry=0$
Rearragnging the terms, we get: $\frac{x^2}{y}+y=2r$
Now differentiating w.r.t. x: $\frac{y\left(2x\right)-x^2{y}^{\prime }}{y^2}+y^{\prime }=0$ i.e., $(x^2-y^2)y^{\prime }=2xy$.