If $r=0$, then $t=0$. Thus, $r\neq 0$ and $$s = \frac{t}{4r\sinh q}.$$ Equation is then equivalent to $$r^2 + \frac{t^2}{16r^2\sinh^2 q} +\frac{Ht}{2} = \frac{1}{4} \iff r^4 - \frac{1-2Ht}{4}r^2 + \frac{t^2}{16\sinh^2q} = 0.$$ Using $\coth q = H$, the discriminant of this quadratic polynomial in $r^2$ is $$\Delta(t) = \frac{1}{16}\left( 1-4Ht+4t^2 \right)$$ which in turn is a quadratic polynomial in $t$ with discriminant $$\widetilde{\Delta} = \frac{H^2-1}{16}>0$$ because $H>1$. Thus, $$\Delta(t) = \frac{(T_1-t)(T_2-t)}{4}$$ because $H=\coth q$. Using , $\Delta(t)\geq 0$ if, and only if $t\leq T_1$ and $$r^2 = \frac{1}{8}\left( 1-2Ht \pm 2\sqrt{(T_1-t)(T_2-t)} \right).$$ By symmetry of Equations and , $s^2$ is as in .

We consider the two continuous parametrisations of $r$ and $s$ for $t\in(-\infty,T_1)$ such that $(r,s)$ satisfies Equations and with $w=2\pi t$: $$\label{eqrst}
    \left\{
    \begin{array}{l}
    r(t) := \frac{\pm 1}{2\sqrt{2}}\left(1-2Ht + 2\sqrt{T_1-t}\sqrt{ T_2-t } \right)^{\frac{1}{2}},\\
    s(t) := \frac{t}{4r(t)\sinh q}.
    \end{array}
    \right.$$

Choosing the parametrisation satisfying $r>s$ maps the unit circle of ${\mathbb{C}}^*$ onto a parallel circle of maximal radius, called a “bulge” of the Delaunay surface. As $t$ tends to $0$, the immersions tend towards a parametrisation of a sphere on every compact subset of ${\mathbb{C}}^*$, which is why we call this family of immersions “the spherical family”. When 