Proximal Operator for $ f \left( x \right) = \sqrt{ {x}^{2} + {a}^{2} } $

Question

I need to find the proximal operator for $f(x) = \lambda \sqrt{x^2 + a^2}$, where $x \in \mathbb{R}$ and $a, \lambda >0$ are some constants. Is there a closed form for it? I tried to derive the operator in a straightforward way, i.e., using its formal definition $${\rm prox}_f(y) = \inf_x \Big\{ \frac{1}{2} | x - y |^2 + \lambda \sqrt{x^2 + a^2} \Big\}$$ This approach, however, leads to a quartic equation, thus necessitating numerical root finding with subsequent deliberations as to which of the four roots to keep. I have also tried to take advantage of Moreau decomposition to express ${\rm prox}_f(y)$ in terms of ${\rm prox}_{f^\ast}(y)$, where $f^\ast$ is the convex conjugate of $f$ given by $$f^\ast(y) = \begin{cases}-a \sqrt{\lambda^2 - y^2}, \quad &|y| \le \lambda \\ \infty, \quad &\mbox{otherwise} \end{cases}$$ (if I am not mistaken). Unfortunately, this does not seem to simplify the original problem either. Finally, I have tried a number of numerical approaches (fixed point, localization, guarded Newton). However, every one of them exhibits (painfully) slow convergence when $a$ is relatively small (e.g., $a = 0.01$). So, I still hope there might be a way to find a closed form expression for ${\rm prox}_f(y)$. Any help in this regard would be greatly appreciated. Thanks!