We can define a closed disk $D$ with center $c$ and radius $r$ as the set of points $x$ satisfying $f(x) \le 1$ where $f(x) = \frac1{r^2}\lVert x-c \rVert^2$.
Now take two disks $D_0,\,D_1$ with associated functions $f_0,\, f_1$. Consider the set of points $x$ satisfying $f_\lambda(x) = (1 - \lambda)f_0(x) + \lambda f_1(x) \le 1$ where $0 \le \lambda \le 1$. It turns out this set of points is either empty or a third disk with radius at most the maximum of the radii of $D_0$ and $D_1$.
This is noted in passing by Emo Welzl in Smallest enclosing disks (balls and ellipsoids) (http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.46.1450), where he says it can be shown using "elementary calculations".
How can we prove this? We can't show $f_\lambda(x) = \frac1{r^2}\lVert x - c \rVert^2$ where $r$ and $c$ are the radius and center of the resulting disk. The right hand side is $0$ at $c$ whereas the left is not in general. Convexity seems important. Beyond that it's not obvious to me how to proceed.
Turns out this was indeed straightforward.
We have $f_\lambda(x) = \alpha[(x-a_0)^2 + (y-b_0)^2] + \beta[(x-a_1)^2 + (y-b_1)^2]$. For $f_\lambda(x) \le 1$ to define a disk (or the empty set) we need only that $f_\lambda(x) = \gamma[(x-a_2)^2 + (y-b_2)^2] + \Delta$ for some $\gamma, a_2, b_2, \Delta$. It is enough to complete the square and note that the coefficients of the $x^2, y^2$ are the same so the coefficients of $(x-a_2)^2, (y-b_2)^2$ will be the same.