The differential equation representing all possible curves that cut each member of the family of circles $x^2+y^2-2 C x=0$ (C is a parameter) at right angle, is

$\frac{d y}{d x}=\frac{2 x y}{x^2-y^2}$

Here, we have to find the orthogonal trajectories of the family of circles

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

Differentiating (i) w.r.t. $x$, we get

$2 x+2 y \frac{d y}{d x}-2 C=0 \Rightarrow C=x+y \frac{d y}{d x}$                 ........(ii)

From (i) and (ii), we obtain

$x^2+y^2-2 x\left(x+y \frac{d y}{d x}\right)=0$            [By eliminating C]

$\Rightarrow y^2-x^2-2 x y \frac{d y}{d x}=0 \Rightarrow y^2-x^2=2 x y \frac{d y}{d x}$            ........(iii)

This is the differential equation representing the given family of circles.

To find the differential equation of the orthogonal trajectories, we replace $\frac{d y}{d x}$ by $-\frac{d x}{d y}$ in equation (iii).

Thus, the differential equation representing the orthogonal trajectories is

$y^2-x^2=-2 x y \frac{d x}{d y}$ or, $\frac{d y}{d x}=\frac{2 x y}{x^2-y^2}$