## Orthogonal Trajectroes of System of Circles Centred on x Axis Passing Through the Origin

Consider the set of circles passing through the origin, with centres on the\[x\]

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 \[b/2\]

.