 Solution

A curve element has arc length $ds=\sqrt{1+z^{\prime 2}}dx$ and mass $dm=\mu ds$. Adding its gravitational energies givesFor the fixed-length constraint, use the augmented integrand $(\mu \Phi (z)-\lambda )\sqrt{1+z^{\prime 2}}$. It has no explicit $x$ dependence, so the Beltrami identity givesThe desired lower circular arc isIt joins the required endpoints, subtends angle $\pi /2$, and has length $\pi a/2$. Its slope gives $\sqrt{1+z^{\prime 2}}=a/(h-z)$. Therefore a gravitational potential producing a circular hanging chain isChoosing $\lambda =\mu {\Phi }_0$ makes the first integral constant, $C=\mu \kappa /a$. Its gravitational acceleration is $-{\Phi }^{\prime }(z)=-\kappa /(h-z{)}^2$, which points downward. An arbitrary additive potential constant changes no force.

This construction really gives a minimum. With $w=h-z>0$, the variable part of $V$ is $\mu \kappa \int ds/w$; the ${\Phi }_0$ term is fixed by the length. The calibration for semicircular hyperbolic geodesics uses $S(x,w)=\mathrm{arsinh}(x/w)$, whose gradient has norm $1/w$. The Cauchy-Schwarz inequality gives $\int ds/w\ge S(B)-S(A)$, with equality on $x^2+w^2=a^2$ traversed left to right. Thus the circular arc minimizes the energy among competitors of the same length and endpoints in the stated potential domain, rather than merely satisfying a necessary stationarity condition.