#### Answer

$\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$

#### Work Step by Step

given that
$x^2+y^2=2cx$
$c=\frac{x^2+y^2}{2x}$
differentiate the above term with respect to x
$2x+2y\frac{dy}{dx}=2c$
$\frac{dy}{dx}=\frac{c-x}{y}$
$\frac{dy}{dx}=\frac{\frac{x^2+y^2}{2x}-x}{y}$
$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$
for orthogonal trajectory
$\frac{dy}{dx}=\frac{-1}{f(x,y)}$
$\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$
thus,the final answer
$\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$