Answer
See below
Work Step by Step
Given: $x^2+y^2=2cx$
Rewrite: $(x-c)^2+y^2=c^2$
Differentiate: $2(x-c)+2y\frac{dy}{dx}=0\\
\rightarrow \frac{dy}{dx}=-\frac{x-c}{y}\\
\rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}$
Let the slope be $m_2$
We have: $\tan (\frac{\pi}{4})=\frac{m^2-\frac{y^2-x^2}{2xy}}{1+m_2.\frac{y^2-x^2}{2xy}}=1\\
\rightarrow m_2=\frac{-x^2+2xy+y^2}{x^2+2xy-y^2}\\
\rightarrow \frac{dy}{dx}=\frac{-x^2+2xy+y^2}{x^2+2xy-y^2}$
If we let $y=vx$, then $v+x\frac{dv}{dx}=-\frac{v^2+2v-1}{v^2-2v-1}\\
\rightarrow \frac{-v^2+2v+1}{v^3-v^2+v-1}=\frac{1}{x}dx$
Integrate: $\int \frac{-v^2+2v+1}{v^3-v^2+v-1}=\int \frac{1}{x}dx\\
\int (\frac{1}{v-1}-\frac{2v}{v^2+1})dv=\int \frac{1}{x}dx\\
\rightarrow \ln (v-1)-\ln(v^2+1)=\ln x+\ln c_1\\
\rightarrow \ln(\frac{v-1}{x(v^2+1)})=\ln c_1$
Substitute back $vx=y \rightarrow \frac{y-x}{x^2+y^2}=c_1$
The family of orthogonal trajectories can be $y=(x-c_2)^2+(y-c_2)^2=2c_2^2$
