Orthogonal trajectories elimination of constant $$x^2+y^2-2ay=0$$
I just learnt orthogonal trajectories today. So basically we need to eliminate the constants and solve $dy/dx=f(x,y)$ right? How do I eliminate $2ay$ in this question.I tried a bunch of stuff but couldn't get it.
This is how to get the differential equation of the family of curves
$$
x^2+y^2-2ay=0 \ \implies\ a=\frac{x^2+y^2}{2y}\\
$$
by diffrentiating the given equation we get
$$
2x +2yy' -2ay' = 0  \Rightarrow y' = \frac{x}{a-y}\\
$$
substituting the value of $a$ in the expression for $y'$ we get
$$
y' = \frac{2yx}{x^2-y^2}\\
$$
The equation is $x^2+ y^2-2ay= 0$.  Differentiate with respect to x:  $2x+ 2yy'-2ay'= 0$. $2(y- a)y'= 2x$, $y'= \frac{x}{y- a}$. Now solve the original equation for a: $2ay= x^2+ y^2$ so $a= \frac{x^2+ y^2}{2y}$.  Replace a in the differential equation with that: $y'= \frac{x}{y+ \frac{x^2+ y^2}{2y}}= \frac{2xy}{2y^2+ x^2+ y^2}= \frac{2xy}{x^2+ 3y^2}$.