We take the Young-Laplace pressure $\sigma \nabla \cdot \mathbf{n}$ due to surface tension
seriously, and consider stationary situations of a fluid in a gravitational
field $-g\mathbf{e}_z$ in which there is an over-pressure $\Delta p$, that is $\sigma \nabla \cdot \mathbf{n} = \Delta p - \rho g z $. Note that this introduces two natural scales, namely the capillary length $l_c\equiv \sqrt{\sigma/\rho g}$ and the capillary pressure $p_c\equiv \sqrt{\sigma \rho g}$ obeying $l_c p_c = \sigma$. In this section, we shall measure lengths in $l_c$ and pressures in $p_c$ unless otherwise stated. In that case, the stationarity condition reads $\Delta p - z = \nabla \cdot \mathbf{n}$, where $\Delta p$ is taken to be constant. For a generic surface $G(\mathbf{r})=0$ we have
\begin{equation}
\nabla \cdot \mathbf{n}
= \nabla \cdot \frac{\nabla G}{|\nabla G|}
= \frac{\nabla^2 G}{|\nabla G|}
-
\frac{\nabla G \cdot \nabla |\nabla G|}{|\nabla G|^2}
\label{eq:divergenceOfNormal}
\end{equation}
which, in the rotationally symmetric (monotonic radial) case $G(r,z)=Z(r)-z$ gives
\begin{equation}
Z''(r) =
\left( \Delta p-Z \right)
\left(1+Z'^2\right)^\frac{3}{2}
- \frac{1}{r}Z' \left(1+Z'^2\right).
\label{eq:DropletInterfaceEquationZ}
\end{equation}
Likewise, for the monotonic vertical case $G(r,z)=r-R(z)$ we get
\begin{equation}
R''(z) = \left( z-\Delta p \right)\left(1+R'^2\right)^\frac{3}{2}
+\frac{1}{R}(1+R'^2)
\label{eq:DropletInterfaceEquationR}
\end{equation}
Notice that a hanging droplet, whose initial conditions can be set to $Z_0'(r)=0$, $Z_0=0$ and $r_0=0$, leaves only one degree of freedom for the shape -- the over-pressure $\Delta p$. In other words, if one were to take an image of a hanging droplet from the side, the surface tension $\sigma$ and the buoyancy $\rho g$ can be found by finding $\Delta p$ such that the solutions to \eqref{eq:DropletInterfaceEquationR} matches the shape of the droplet silhouette.
Although exact solutions of \eqref{eq:DropletInterfaceEquationR} and \eqref{eq:DropletInterfaceEquationZ} may be found, to understand the general shape of these interfaces, it can be beneficial to recognise the two principal radii of curvature $R_1$ and $R_2$ defined by $\Delta p - z = 1/R_1 + 1/R_2$. Indeed, we may identify these radii in equations \eqref{eq:DropletInterfaceEquationR} and \eqref{eq:DropletInterfaceEquationZ}:
\begin{align*}
\Delta p - z &= \frac{1}{R_1} + \frac{1}{R_2}
\text{ for } R_1 \equiv -\frac{\left(1+R'^2\right)^\frac{3}{2}}{R''}
\text{ and } R_2 \equiv \sqrt{1+R'^2} R
\\
\Delta p-Z &= \frac{1}{R_1} + \frac{1}{R_2}
\text{ for } R_1 \equiv \frac{\left(1+Z'^2\right)^\frac{3}{2}}{Z''}
\text{ and }
R_2 \equiv \frac{r\sqrt{1+Z'^2}}{Z'}.
\end{align*}
At the bottom of a droplet, where $R,Z,Z' \simeq 0$, rotational symmetry requires that we have $R_1=R_2=R_c$. Then $R_c=2/\Delta p$. Notice that this case also captures the radii of spherical droplets with over-pressure $\Delta p$.
Note also that the radius can at most be $R_{max} = 1/|\Delta p-z|$,and if $2\pi R < Z$ we encounter the Plateau–Rayleigh instability (rough estimate). So for a stable droplet, we need $R\in[R_{min},R_{max}]$ for all $R$, with $R_{min}(z) = z/2\pi$ and $R_{max}(z)=1/|\Delta p-z|$.