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