Derivative of circumradius Let $P = \{p_1, p_2, \cdots, p_n\}\subset\mathbb{R}^2 - \mathbb{R}\times\{0\}$ be a set of $n$ distinct points. Define the function $f: \mathbb{R}\times\{0\}\to\mathbb{R}^+$ by
$$(t,0)\mapsto \text{radius of the smallest enclosing circle of $P\cup\{(t,0)\}$}$$
Is this function of class $C^1$? Note that the metric is the Euclidean metric. 
 A: Extending my comment, the "circumradius" is not $C^1$. In general, it is piecewise $C^{\infty}$. In fact, its square is a piecewise rational function in $t$.
For any finite set of points $X = \{x_1, x_2, \ldots, x_n\} \subset \mathbb{R}^2$, let $\mathscr{R}(X)$ be the circumradius of $X$. ie. the minimal radius of circle containing all points in $X$.
In particular, for singletons $\{a\}$ or pair of points $\{a,b\}$, we have/define:
$$\mathscr{R}(\{a\}) = 0\quad\text{ and }\quad\mathscr{R}(\{a,b\}) = \frac{|a-b|}{2}$$
It is easy to verify the circumradius of $X$ is determined by the circumradius of its subsets of $3$ or fewer elements:
$$\mathscr{R}(X) = \max_{(a,b,c) \in X^3}\mathscr{R}(\{a,b,c\})\tag{*}$$
To understand how $\mathscr{R}(X)$ behaves, it just suffices to look at
the dependence of $\mathscr{R}(\{a,b,c\})$ upon $a,b,c$. Let us consider
the special case $a = (0,1), b = (0,-1)$ and $c = (x,y)$ where $x, y \ge 0$.
There are 3 possible subcases:


*

*$x^2 + y^2 \le 1$, $c$ falls inside the circle defined by $a,b$ and 
$$\mathscr{R}(\{a,b,c\}) = \mathscr{R}(\{a,b\}) = 1.$$

*$x^2 + y^2 > 1$ and $y \ge 1$, $a$ falls inside the circle defined by $b,c$ and $$\mathscr{R}(\{a,b,c\}) = \mathscr{R}(\{b,c\}) = \frac12\sqrt{x^2 + (y+1)^2}.$$

*$x^2 + y^2 > 1$ and $0 \le y < 1$, with a little bit of geometry, one can show
$$\mathscr{R}(\{a,b,c\}) = \sqrt{1 + \left(\frac{x^2 + y^2 - 1}{2 x}\right)^2}$$
This means if we fix $a,b$ and let $c$ moves along a straight line $c(t) = (x_0,y_0) + t (x',y')$,
$\mathscr{R}(\{a,b,c(t)\})$ will be a function composed of pieces of 3 possible
forms:


*

*a constant.

*$\sqrt{P_2(t)}$ - square root of a quadratic function in $t$.

*$\frac{\sqrt{P_4(t)}}{P_1(t)}$ - square root of a $4$-th degree polynomial divided by a linear polynomial.


In short, $\mathscr{R}(\{a,b,c(t)\})^2$ will be a piecewise rational function in $t$.
Back to the original circumradius we want to study. Using $(*)$, we have:
$$\mathscr{R}(P\cup\{(t,0)\}) = \max\left(\mathscr{R}(P), \max_{(a,b)\in P^2}(\mathscr{R}(\{a,b,(t,0)\})\right)$$
This implies the square of the circumradius we want to calculate is the maximum of a bunch of piecewise rational functions in $t$. 
Given any 2 rational functions $f(t), g(t)$, we know either $f(t) = g(t)$ for all $t$ or for finite many $t$. This implies the maximum of 2 piecewise rational functions is still a piecewise rational function.
From this, we can conclude $\mathscr{R}(P\cup\{(t,0)\})^2$ is a piecewise rational function and hence the circumradius we care, $\mathscr{R}(P\cup\{(t,0)\})$, is piecewise $C^{\infty}$.
UPDATE 
About the assertion that the "circumradius" $r(t) = \mathscr{R}( P \cup \{ (t,0)\} )$ is not $C^1$. The counter-example I used before:
$$P = \{ (0,1), (0,-1) \}$$
didn't work. I failed to recognize in that case, 
$$\lim_{t\to 1+} r'(t) = \lim_{t\to 1-} r'(t) = 0$$
In fact, for any 2 points $a, b$ and straight line $c(t) = (x_0,y_0) + t (x',y')$, The function $\mathscr{R}(\{a,b,c(t)\})$ is $C^1$ over $\mathbb{R}$ as long as  $a \ne b$ and $c(t)$ didn't intersect $a$ nor $b$. 
This implies when $|P| = 2$, $r(t)$ is indeed $C^1$ over $\mathbb{R}$.
When $|P| = 3$, there are already examples that $r(t)$ fails to be $C^1$. However, those examples are hard to visualize. The simplest example I find is for $|P| = 4$ and the points
forms the corner of a square: 
$$P = \{ (1,1), (1,-1), (-1,1), (-1,-1) \}$$
With a little bit of algebra, one can show:
$$r(t) = \begin{cases}
    \frac{(|t|+1)^2+1}{2(|t|+1)},&|t| \ge \sqrt{2}\\
    \sqrt{2}, & |t| \le \sqrt{2}
\end{cases}$$
Notice in this case,
$$\lim_{|t|\to\sqrt{2}+} |r'(t)| = \sqrt{2} - 1 \ne 0 = \lim_{|t|\to\sqrt{2}-} |r'(t)|$$
This means $r(t)$ fails to be differentiable at $t = \pm\sqrt{2}$ and hence is not $C^1$.
