The orthogonal trajectories of the family of circles given by $x^2+y^2-2ay=0$ (a is the parameter), is

The equation of the family of circles is

$x^2+y^2-2ay=0$                                   ... (i)

Differentiating w.r.t. x, we get

$2x+2y\frac{dy}{dx}-2a\frac{dy}{dx}=0\Rightarrow a=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}$

Putting this value of a in (i), we get

$x^2+y^2-2y\left\{\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}\right\}=0\Rightarrow x^2-y^2-2xy\frac{dx}{dy}=0$        ... (ii)

This is the differential equation of the family of circles given by (i). The differential equation representing the family of orthogonal trajectories of (i) is obtained by replacing $\frac{dy}{dx}by-\frac{dx}{dy}$ in (ii). So, the differential equation of the orthogonal trajectories is.

$\begin{array}{l}{x}^2-y^2+2xy\frac{dy}{dx}=0\\ \Rightarrow \left(x^2-y^2\right)dx+2xydy=0\\ \Rightarrow 2xydy-y^{2}dx=-x^{2}dx\\ \Rightarrow \frac{xd\left(y^2\right)-y^{2}dx}{x^2}=-dx\\ \Rightarrow d\left(\frac{y^2}{x}\right)=-dx\end{array}$

$\Rightarrow \frac{y^2}{x}=-x+2k\Rightarrow x^2+y^2-2kx=0$

This is the required family of orthogonal trajectories.