Given a uniquely geodesic metric space $X$, let $\mathcal K(X)$ denote the metric space of compact, convex subsets of $X$ equipped with the Hausdorff distance. Given $K \in \mathcal K(X)$, let $c(K)$ be the circumcenter of $K$, i.e. the center of the smallest ball containing $K$ (we will only consider $X$ where $c$ is well-defined).
For $X = \mathbb R^2$, $c$ is not Lipschitz. This can be seen by taking $K$ to be the triangle with vertices $(-1, 0)$, $(1, 0)$, $(1, \epsilon)$, and $K'$ to be the triangle with vertices $(-1, 0), (1, 0), (\sqrt{1-\epsilon^2}, \epsilon)$. Since $K$ and $K'$ are right triangles, their circumcenters are the midpoints of their respective hypotenuses, so $c(K) = (0, \epsilon / 2)$ and $c(K') = (0, 0)$. But $d_H(K, K') \leq 1 - \sqrt{1 - \epsilon^2} \sim \epsilon^2 / 2$.
On the other hand, if $X$ is a metric tree (i.e. a simply connected one-dimensional simplicial complex), it is not hard to show that $c$ is Lipschitz. One can reduce to the case where $K \subset K'$, and use the fact that $c(K)$ is the midpoint of a diameter chord of $K$ to show that $d(c(K), c(K')) \leq \frac{1}{2}(\text{diam}(K') - \text{diam}(K))$.
If $X$ is hyperbolic space $\mathbb H^n$, the first example shows that $c$ is not Lipschitz. But the second example makes me wonder if we can show that if there exist constants $C$ and $L$ such that if $\text{diam}(K), \text{diam}(K') \geq C$, $d(c(K), c(K')) \leq L d_H(K, K')$.
