Orthogonal Trajectroes of System of Circles Centred on x Axis Passing Through the Origin

Consider the set of circles passing through the origin, with centres on the  
  axis. The equations have equations of the form  
  is the radius.
Differentiating and simplifying gives  
\[x-a+y \frac{dy}{dx}=0 \rightarrow a=x+ y \frac{dy}{dx}\]
Substitute this expression for  
  into the equation of the circle and simplify to give  
\[x^2+y^2=2x(x+y \frac{dy}{dx} ) \rightarrow \frac{dy}{dx} = \frac{y^2-x^2}{2xy}\]
. The perpendiculars to the circle at every point on the circle have gradient  
\[-1/ \frac{dy}{dx} = \frac{2xy}{x^2-y^2}\]
This perpendicular gradient function can be written as  
. This equation is homogeneous so substitute  
\[x=vy \rightarrow \frac{dx}{dy}=v+y \frac{dv}{dy} \rightarrow v+ y \frac{dv}{dy}= \frac{y^2v^2-y^2}{2vyy}= \frac{v^2-1}{2v} \]
A little rearrangement followed by separation of variables gives  
\[\frac{2v}{v^2+1} dv + \frac{1}{y} dy\]
Integration gives  
\[ln(1+v^2)+ln(y)=C \rightarrow ln((1+x^2/y^2)y)=C \rightarrow x^2+y^2=by\]
We can rearrange this equation as  
  which is the equation of a circle on with centre at  
, radius  

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