Family of Curves Intersecting Circle Centre the Origin at 45 Degrees

What is the set of curves that intersect the set of circles centred at the origin with equations  
\[x^2+y^2=c\]
  at an angle of  
\[\pi /4\]
?
Tangents to the circle at points  
\[(x,y)\]
  have gradient function  
\[\frac{dy}{dx}=- \frac{x}{y} tan \theta \]
, where  
\[\theta\]
  is the angle the tangent drawn at the point makes with the  
\[x\]
  axis.
Then at the same point, where the curve  
\[C\]
  intersects the circle at an angle of  
\[\pi /4\]
  the tangent to  
\[C\]
  makes an angle  
\[\theta + \frac{\pi}{4}\]
  with the  
\[x\]
  axis.
\[\begin{equation} \begin{aligned} tan( \theta + \frac{\pi}{4}) &= \frac{tan \theta + tan \pi /4}{1- tan \theta tan \pi /4} = \frac{tan \theta +1}{1-tan \theta} \\ &= \frac{-x/y+1}{1- (-x/y)}= \frac{-x+y}{y+x} \end{aligned} \end{equation}\]

This equation is homogeneous so let  
\[y=vx\]
  then  
\[\frac{dy}{dx}=v+x \frac{dv}{dx}\]
. The equation becomes
\[v+x \frac{dv}{dx}= \frac{-x+vx}{vx+x}= \frac{-1+v}{v+1}\]
.
Subtracting  
\[v\]
  gives  
\[x \frac{dv}{dx} = \frac{-1+v}{v+1}-v= -\frac{v^2+1}{v+1}\]
.
Separating variables gives  
\[\frac{v+1}{v^2+1} dv= - \frac{1}{x}dx\]
.
Integration gives  
\[\frac{1}{2} ln(v^2+1)+ tan^{-1}(v)= -ln (x)+A\]
.
Substitute  
\[v=y/x\]
  then  
\[\frac{1}{2} ln(y^2/x^2+1)+ tan^{-1}(y/x)= -ln (x)+A\]
.

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