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

$x^2+y^2=C x$

We have,

$x^2+y^2-a y=0$           ....(i)

$\Rightarrow 2 x+2 y \frac{d y}{d x}-a \frac{d y}{d x}=0 \Rightarrow a=\frac{2\left(x+y \frac{d y}{d x}\right)}{\frac{d y}{d x}}$

Substituting the value of a in (i), we obtain

$x^2+y^2-2 y\left(\frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}}\right)=0 \Rightarrow\left(x^2-y^2\right) \frac{d y}{d x}-2 x y=0$

This is the differential equation of the family of circles given in (i). The differential equation representing the orthogonal trajectories is

$-\left(x^2-y^2\right) \frac{d x}{d y}-2 x y=0$

$\Rightarrow \frac{d y}{d x}=-\left(\frac{x^2-y^2}{2 x y}\right) $

$\Rightarrow 2 x y d y-y^2 d x=-x^2 d x$

$\Rightarrow \frac{x d\left(y^2\right)-y^2 d x}{x^2}=-d x \Rightarrow d\left(\frac{y^2}{x}\right)=-d x$

On integrating, we get

$\frac{y^2}{x}=-x+C \Rightarrow y^2+x^2=C x$