## 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  {jatex options:inline}x^2+y^2=c{/jatex}  at an angle of  {jatex options:inline}\pi /4{/jatex}?
Tangents to the circle at points  {jatex options:inline}(x,y){/jatex}  have gradient function  {jatex options:inline}\frac{dy}{dx}=- \frac{x}{y} tan \theta {/jatex}, where  {jatex options:inline}\theta{/jatex}  is the angle the tangent drawn at the point makes with the  {jatex options:inline}x{/jatex}  axis.
Then at the same point, where the curve  {jatex options:inline}C{/jatex}  intersects the circle at an angle of  {jatex options:inline}\pi /4{/jatex}  the tangent to  {jatex options:inline}C{/jatex}  makes an angle  {jatex options:inline}\theta + \frac{\pi}{4}{/jatex}  with the  {jatex options:inline}x{/jatex}  axis.
{jatex options:inline}\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}{/jatex}
This equation is homogeneous so let  {jatex options:inline}y=vx{/jatex}  then  {jatex options:inline}\frac{dy}{dx}=v+x \frac{dv}{dx}{/jatex}. The equation becomes
{jatex options:inline}v+x \frac{dv}{dx}= \frac{-x+vx}{vx+x}= \frac{-1+v}{v+1}{/jatex}.
Subtracting  {jatex options:inline}v{/jatex}  gives  {jatex options:inline}x \frac{dv}{dx} = \frac{-1+v}{v+1}-v= -\frac{v^2+1}{v+1}{/jatex}.
Separating variables gives  {jatex options:inline}\frac{v+1}{v^2+1} dv= - \frac{1}{x}dx{/jatex}.
Integration gives  {jatex options:inline}\frac{1}{2} ln(v^2+1)+ tan^{-1}(v)= -ln (x)+A{/jatex}.
Substitute  {jatex options:inline}v=y/x{/jatex}  then  {jatex options:inline}\frac{1}{2} ln(y^2/x^2+1)+ tan^{-1}(y/x)= -ln (x)+A{/jatex}.