The physics dictating the shape at the bottom is subtle, but assume for simplicity that the shape is a bit more than a semicircle on top, and a parabola at the bottom.
Then if we take the center of the circle to be the origin, there are two parameters to describe the balloon: The angle past a semicircle that is still in the "upper" part, and the circle radius. I will take the radius to be one; I assume you can scale for a larger balloon.
The angle $\theta$ beyond a semicircle is a bit arbitrary but it looks from the picture that it it goes $30^\circ$ past the semicircle. Then The parabola
goes thru the points $(-\frac{\sqrt{3}}{2}.-\frac12)$ and $(+\frac{\sqrt{3}}{2}.-\frac12)$, and has slope $\sqrt{3}$ at $x=\frac{\sqrt{3}}{2}$. This specifies the parabola unicles, so the shape of the balloon is
$$
y = \left\{ \begin{array}{ccccl}
+\sqrt{1-x^2} &,& -\sqrt{1-x^2} & \mbox{when }& -1 \leq x \leq -\frac{\sqrt{3}}{2} \\
+\sqrt{1-x^2} &,& \frac{\sqrt{3}}{2} x^2 - \frac{4+3\sqrt{3}}{8} & \mbox{when }&
-\frac{\sqrt{3}}{2} < x < \frac{\sqrt{3}}{2} \\
+\sqrt{1-x^2} &,& -\sqrt{1-x^2} &\mbox{when }& x \leq +\frac{\sqrt{3}}{2} \leq 1
\end{array} \right.
$$