\[x\]
axis. The equations have equations of the form \[(x-a)^2+y^2=a^2\]
where \[a\]
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
\[a\]
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
\[\frac{dx}{dy}=\frac{x^2-y^2}{2xy}\]
. 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
\[ln(1+v^2)+ln(y)=C \rightarrow ln((1+x^2/y^2)y)=C \rightarrow x^2+y^2=by\]
where \[b=e^C\]
.We can rearrange this equation as
\[x^2+(y-b/2)^2=b^2/4\]
which is the equation of a circle on with centre at \[y=b/2\]
, radius \[b/2\]
.