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For uniform linear mass density $\mu$, a chain minimizing $\mu \int \Phi (z)ds$ at fixed length has Beltrami identity $(\mu \Phi -\lambda )/\sqrt{1+z^{\prime 2}}=C$. In the domain $z<h$, the displayed potential has downward gravitational acceleration $-{\Phi }^{\prime }(z)=-\kappa /(h-z{)}^2$. A circular arc $z=h-\sqrt{a^2-x^2}$ satisfies the identity because $\sqrt{1+z^{\prime 2}}=a/(h-z)$.

This arc is a global minimum among admissible competitors of the same length with the same endpoints in $z<h$. Set $w=h-z>0$. The variable part of the energy is $\mu \kappa \int ds/w$, a hyperbolic length in the Poincaré half-plane model. The calibration for semicircular hyperbolic geodesics $S=\mathrm{arsinh}(x/w)$ has $|\nabla S|=1/w$. The Cauchy-Schwarz inequality gives $\int ds/w\ge S(B)-S(A)$, with equality along the circle traversed with increasing $x$. The constant-potential part is unchanged by the fixed-length constraint.

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.

Interpret the density $1/y$ on the physical region $y>0$. It is singular at zero and would not describe a nonnegative density for $y<0$. The first integral gives $y\sqrt{1+y^{\prime 2}}=R$ for a positive constant $R$. To handle the horizontal tangent without dividing by $y^{\prime }$, use the full Euler-Lagrange equation:Two integrations give $y^2=-(x-x_0{)}^2+R^2$. The equal endpoint heights force $x_0=0$, and either endpoint then gives $R^2=2$. ThusThe path is the upper circular arc. A constant horizontal path solves the differentiated first integral trivially but fails $y{y}^{\prime \prime }+y^{\prime 2}+1=0$, so is not another extremum.

One can prove global minimality, rather than only stationarity. Define $\Phi (x,y)=\mathrm{arsinh}(x/y)$ on the upper half-plane. ThenThe Cauchy-Schwarz inequality gives, along any admissible path from $A$ to $B$,Our circle has tangent $(1,-x/y)$ for increasing $x$, so equality holds pointwise. This calibration for semicircular hyperbolic geodesics provesThe geometry is that of a hyperbolic geodesic in the Poincaré half-plane model. The proof even allows competing upper-half-plane paths beyond the class of monotonic graphs.