Show that the path of shortest hyperbolic length satisfies $(x-c)^2+y^2=r^2$

Question

The hyperbolic length of a curve $y:[a,b]\rightarrow\mathbb{RxR}_+$ is given by the functional $$\lambda(y)=\int_a^b\frac{\sqrt{1+y'^2}}{y}dx$$ Show that the path of shortest hyperbolic length satisfies $(x-c)^2+y^2=r^2$ for $c,r\in\mathbb{R}$. We are approaching this problem using calculus of variations and I've used the Euler-Lagrange differential equation to show: $$\frac{d}{dx}(\frac{y'}{y\sqrt{1+y'^2}})=\frac{-\sqrt{1+y'^2}}{y^2}$$ and $$\frac{1}{y\sqrt{1+y'^2}}=C$$ where C is a constant, but now I've been stuck for a while. Does anyone have advice or a hint they could give me?

Answer 1

With $c=1/C^2$ your ODE is $y^2(1+y'^2)=c$, which solved for $yy'$ is $$yy'=\sqrt{c-y^2}.$$ Now put $u=y^2$ so that $u'=2yy',$ and get to $$\frac{u'}{\sqrt{c-u}}=2.$$ With $v=c-u$ and $dv=-du$ integrating the sides gives $$v^{1/2}=-x+k/2,$$ so $v=(x-k/2)^2$, then $c-u=c-y^2=(x-k/2)^2$, i.e. $$(x-k/2)^2+y^2=c.$$