## 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
$(x,y)$
$\frac{dy}{dx}=- \frac{x}{y} tan \theta$
, 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{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}

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
$\frac{1}{2} ln(y^2/x^2+1)+ tan^{-1}(y/x)= -ln (x)+A$
.