## 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
$a$
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
$y=b/2$
$b/2$ 