Analytic continuation of $g(\sqrt{1-x^2})$ where $g$ is entire

Question

Consider a function $f(x)=\sqrt{1-x^2}, x\in \mathbb{R}$. Moreover, let $g:\mathbb{C} \to \mathbb{C}$ be an entire function and consider the composition \begin{align} h(x)= g(f(x))=g(\sqrt{1-x^2}) \end{align}

Question: Let $H(z)$ be the complex analytic extension of $h(x)$ and let $U$ denote the domain of $H(z)$. What is $U$?

Some thoughts

1. I think, since, $g$ is entire we can ignore it. What I mean is that $U$ will be the domain to which the function $f(x)=\sqrt{1-x^2}$ can be extended. Therefore, we need to find a domain of continuation for $f(x)$. (As phrased this technically wrong. See the comment below for a correct statement).
2. I think the candidate for the extension is $F(z)=\sqrt{1-z^2}, z\in \mathbb{C}$ since $f(x)=F(x), x \in [-1,1]$.
3. The function $f(x)$ has a power series expansion around zero with a radius of convergence $1/2$. Therefore, it must have analytic continuation at least to $|z|<1/2$.

Edit Based on the comments and suggestions. Let's restrict our attention only to the principle branch of square root.