Find the orthogonal trajectories of the circles  $x^2+y^2-ay=0,$  where $a$ is a parameter.

Given curve is

$\begin{array}{c}{x}^2+y^2-ay=0\\ 2x+2y\frac{dy}{dx}-a\frac{dy}{dx}=\\ \Rightarrow 2x+2y\frac{dy}{dx}=0\\ a=\frac{dy}{dx}\end{array}$

Eliminating a between (i) and (ii), we get

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(x^2+y^2\right)-\left(\frac{2x+2y\frac{dy}{dx}}{\frac{dy}{dx}}\right)y=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(x^2+y^2\right)\frac{dy}{dx}-\left(2x+2y\frac{dy}{dx}\right)y=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(x^2+y^2-2y^2\right)\frac{dy}{dx}=2xy\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(x^2-y^2\right)\frac{dy}{dx}=2xy\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{dy}{dx}=\frac{2xy}{\left(x^2-y^2\right)}\end{array}$

Replace $dy/dx\text{ by }-dx/dy$, we get

$\begin{array}{l}-\frac{dx}{dy}=\frac{2xy}{\left(x^2-y^2\right)}\\ \Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}\\ =\frac{{\left(\frac{y}{x}\right)}^2-1}{2\left(\frac{y}{x}\right)}\\ \Rightarrow \frac{v+x\frac{dv}{dx}=\frac{v^2-1}{2v}}{,}v=\frac{y}{x}\\ x\frac{dv}{dx}=\frac{v^2-1}{2v}-v\\ =-\frac{v^2-1}{2v}\\ \frac{2vdv}{\left(v^2+1\right)}=-\frac{dx}{x}\end{array}$

Integrating, we get

$\log \left|v^2+1\right|+\log \left|x\right|=\log c$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x\left(v^2+1\right)=c\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}^2+y^2-cx=0\end{array}$

Which is the required orthogonal trajectories.