Fitting a closed curve on the roots of ${x \choose k}-c$ Let
$${n \choose k} = \frac1{(n+1) \operatorname{B}(n-k+1, k+1)}.$$
be the generalized binomial coefficient, and here $\operatorname{B}$ is the Beta function.
Let $f_{k,c}(x)$ be the following function
$$f_{k,c}(x) = {x \choose k}-c,$$
where $k$ and $c$ are nonzero scalars.
Are there a special general closed curve, what we can fit on the roots of $f_{k,c}$?
The roots on the complex plane for $(k,c)=(6,1)$, $(k,c)=(9,1)$, $(k,c)=(15,1)$ are the followings.

The results are similar for other $(k,c)$ pairs. First I thought that it is an ellipse, but actually after fitting ellipses on arbirtrary five roots of higher order $f_{k,c}$ functions it turned out that it is not. The motivation of the problem was this question.
 A: For convenience let me set $k = 2n$.
We can represent the binomial coefficient by
$$
\binom{x}{2n} = \frac{1}{(2n)!}\prod_{j=0}^{2n-1} (x-j),
$$
so to make this an even polynomial we will replace $x$ by $y+n-\tfrac{1}{2}$.  This will make the roots of the equation
$$
\binom{y+n-\tfrac{1}{2}}{2n} = c \tag{0}
$$
symmetric about the real and imaginary axes (at least when $c$ is real), effectively centering the "oval" at the origin.
It appears that the roots tend to $\infty$ as $n \to \infty$.  From the product representation we observe that
$$
\binom{y+n-\tfrac{1}{2}}{2n} \sim \frac{y^{2n}}{(2n)!}
$$
for large $y$, so we might expect that equation $(0)$ behaves a little like
$$
\frac{y^{2n}}{(2n)!} \approx c
$$
when $n$ is large.  Multiplying both sides by $(2n)!$ and raising things to the $1/(2n)$ power the "equation" becomes, by Stirling's formula,
$$
y \approx \frac{2n}{e}.
$$
This heuristic reasoning leads us to suspect that the roots grow approximately on the order of $n$.  We'll make one more substitution to reflect this, replacing $y$ by $nz$.
In total we want to make the substitution
$$
x = nz + n - \tfrac{1}{2}
$$
into the original equation $\binom{x}{2n} = c$ and hence study the new equation
$$
\binom{nz + n - \tfrac{1}{2}}{2n} = c. \tag{1}
$$
Using the Beta function representation of the binomial coefficient we can rewrite it in terms of Gamma functions as
$$
\binom{nz + n - \tfrac{1}{2}}{2n} = \frac{\Gamma\left(nz+n+\tfrac{3}{2}\right)}{\left(nz+n+\tfrac{1}{2}\right) \Gamma\left(nz-n+\tfrac{1}{2}\right) \Gamma(2n+1)}.
$$
We now replace the Gamma functions with their Stirling formula equivalents;
$$
\Gamma\left(nz+n+\tfrac{3}{2}\right) \sim \left(\frac{nz+n+\tfrac{3}{2}}{e}\right)^{nz+n+3/2}\sqrt{\frac{2\pi}{nz+n+\tfrac{3}{2}}},
$$
$$
\Gamma\left(nz-n+\tfrac{1}{2}\right) \sim \left(\frac{nz-n+\tfrac{1}{2}}{e}\right)^{nz-n+1/2}\sqrt{\frac{2\pi}{nz-n+\tfrac{1}{2}}},
$$
$$
\Gamma(2n+1) \sim \left(\frac{2n+1}{e}\right)^{2n+1} \sqrt{4\pi n},
$$
where the first two are valid for $\operatorname{Im} z \neq 0$ with a relative error of $O(1/n)$ as long as $z$ remains bounded away from the points $z=\pm 1$.  With a little bit of algebra we find that
$$
\binom{nz + n - \tfrac{1}{2}}{2n} \sim \frac{(2n+1)^{-2n-1}}{\sqrt{4\pi n} \left(nz+n+\tfrac{1}{2}\right)} \sqrt{\frac{nz-n+\tfrac{1}{2}}{nz+n+\tfrac{3}{2}}} \frac{\left(nz+n+\tfrac{3}{2}\right)^{nz+n+3/2}}{\left(nz-n+\tfrac{1}{2}\right)^{nz-n+1/2}}.
$$
A straightforward calculation will show that
$$
\frac{\left(nz+n+\tfrac{3}{2}\right)^{nz+n+3/2}}{\left(nz-n+\tfrac{1}{2}\right)^{nz-n+1/2}} \sim e n^{2n+1} \frac{(z+1)^{nz+n+3/2}}{(z-1)^{nz-n+1/2}}
$$
as $n \to \infty$, so on taking absolute values and raising both sides to the $1/n$ power we obtain
$$
\left|\binom{nz + n - \tfrac{1}{2}}{2n}\right|^{1/n} = \frac{1}{4} \left|\frac{(z+1)^{z+1}}{(z-1)^{z-1}}\right|\left(1+O\left(\frac{\log n}{n}\right)\right),
$$
as $n \to \infty$, where the error term holds uniformly with respect to $z$ as long as $z$ remains bounded away from the points $z = \pm 1$.  So, after taking absolute values of both sides of $(1)$ followed by $n^\text{th}$ roots,
$$
\left|\binom{nz + n - \tfrac{1}{2}}{2n}\right|^{1/n} = |c|^{1/n},
$$
letting $n \to \infty$ yields
$$
\frac{1}{4} \left|\frac{(z+1)^{z+1}}{(z-1)^{z-1}}\right| = 1.
$$
So...

The roots of equation $(1)$ tend to the limit curve
  $$
\left|\frac{(z+1)^{z+1}}{(z-1)^{z-1}}\right| = 4 \tag{2}
$$
  as $n \to \infty$.

Below is a plot of the roots of equation $(1)$ with $n=30$ and $c=1$ in $\color{blue}{\text{blue}}$ along with the limit curve $(2)$ in $\color{red}{\text{red}}$.

And here is a plot with $n=100$ and $c=1$.

Remark 1: These calculations indicate that the rate at which the zeros approach the limit curve is $O(\log n/n)$ away from the points $z = \pm 1$.  Near these points the rate of approach should be different and a more detailed analysis is required to determine it.
Remark 2: The variable $c$ does not need to remain fixed; it may in fact depend on $n$.  As long as
$$
|c|^{1/n} \to 1
$$
as $n \to \infty$, the roots of equation $(1)$ will still tend to the limit curve $(2)$, though perhaps at a different rate than if $c$ were held fixed.
Remark 3:  With more work it is possible to obtain better approximations to the actual curve on which the roots lie.  Some rough calculations seem to indicate that
$$
\left|\frac{(z+1)^{z+1}}{(z-1)^{z-1}}\right| = 4 + \frac{\log(16e^4|c|^4n^2)}{n}
$$
is a much better one.  Here's a plot with $n = 30$ and $c=5$, showing this 'adjusted' curve as a thin blue line.

