By rolling a circle around an ellipse and tracing the circle's center, one obtains an equidistant curve to the ellipse. I am interested in the polar form of such a curve. Let an ellipse be given by the parametric equation \begin{equation} e(\alpha) = \begin{bmatrix}a\cos\alpha\\ b\sin\alpha \end{bmatrix}, \text{ for } \alpha \in [0, 2\pi), \end{equation}
where $a, b$ are the lengths of the semi-major and semi-minor axis, respectively. Suppose a circle of radius $R$ is rolled along the ellipse $e$, then calculating the normal vector of length $R$ at each point along the ellipse gives an equidistant curve to it, i.e.,
\begin{equation}\label{eq_ellipse_enlargment} t(\alpha) = \begin{bmatrix} a\cos\alpha\\ b\sin\alpha \end{bmatrix} + \frac{R}{\sqrt{b^2\cos^2\alpha + a^2 \sin^2\alpha}}\begin{bmatrix} b\cos\alpha\\ a\sin\alpha \end{bmatrix}. \end{equation}
As the equidistant curve is symmetric w.r.t. to both of its axes, it should be possible to derive a polar form, $r(\phi)$, where $r$ is the radial distance and $\phi$ the polar angle. I tired eliminating the parameter $\alpha$ by the identity $\cos^2\alpha + \sin^2\alpha = 1$, however, ended up nowhere. I also thought about starting with the polar or canonical ellipse representation to derive the equidistant curve, but I also did not go anywhere. I would appreciate any helpful insights; please mind that I am not a mathematician.
Edit: Previously, I thought the equidistant curve is an elliptic epitrochoid, which according to the comments is not the case. I changed the question accordingly.
I would try to obtain an implicit definition first, using algebraic equations, then use that to derive a polar form by looking at the equation of the radius for a given angle.
First let's get rid of those triginometric functions using the tangent half-angle formulas:
$$ t := \tan\frac\alpha2 \qquad \cos\alpha = \frac{1-t^2}{1+t^2} \qquad \sin\alpha = \frac{2t}{1+t^2} $$
Then the point on the ellipse becomes
$$ \begin{bmatrix}a\cos\alpha\\b\sin\alpha\end{bmatrix}= \frac1{1+t^2}\begin{bmatrix}a(1-t^2)\\b\cdot2t\end{bmatrix} $$
and the direction of the offset becomes
$$ \begin{bmatrix}b\cos\alpha\\a\sin\alpha\end{bmatrix}= \frac1{1+t^2}\begin{bmatrix}b(1-t^2)\\a\cdot2t\end{bmatrix} $$
The magnitude of that vector doesn't matter because you need to normalize it anyway, so you might as well ignore the scalar factor in front. The square root from the norm of the remaining vector is a bit of an annoyance. Let's introduce a symbol for that and see how it relates to the square root you had.
$$ s := \sqrt{b^2(1-t^2)^2+4a^2t^2} \qquad \sqrt{b^2\cos^2\alpha+a^2\sin^2\alpha} = \frac{s}{1+t^2} $$
With that you have the point on the offset curve as
$$ \begin{bmatrix}x\\y\end{bmatrix}= \frac1{1+t^2}\begin{bmatrix}a(1-t^2)\\b\cdot2t\end{bmatrix}+ \frac Rs\begin{bmatrix}b(1-t^2)\\a\cdot2t\end{bmatrix} $$
At this point you can rephrase everything as three polynomial equations, one each for the equations of $x,y,s$ and each multiplied by the common denominator to remove the divisions:
\begin{align*} sa(1-t^2) + (1+t^2)Rb(1-t^2) - s(1+t^2)x &= 0 \\ sb\cdot2t + (1+t^2)Ra\cdot2t - s(1+t^2)y &= 0 \\ b^2(1-t^2)^2 + 4a^2t^2 - s^2 &= 0 \end{align*}
Now you can start an exercise in elimination theory. Use the last equation to eliminate $s$ from the other two, then combine the two resulting equations to eliminate $t$. Personally I prefer to use resultants for this, and I use Sage as my computer algebra system of choice to avoid doing all of this manually.
The final resultant has a few short factors that correspond to non-degeneracy conditions: $(a-b)^6(a+b)^6R^8a^8b^8$. The remaining factor is the actual implicit form of the curve we're after. And it's a beast, too big to print here. It contains $x$ and $y$ of combined degree $8$, so it's an octic curve at first glance. Looking more closely, all the degrees are even so you might also see it as a quartic curve in $x^2$ and $y^2$.
Now if you plug $x=r\cos\theta$ and $y=r\sin\theta$ into this equation, you get a quartic equation in $r^2$. And a quartic equation can be solved in terms of radicals, even if the resulting expressions are far from pretty.
So including complex solutions, you would expect to in general get 4 solutions for $r^2$. You would only care about those that are real and positive, and then use $r$ as the positive root of that. This should still leave you with two choices, one for the circle rolling on the outside of the ellipse and one on the inside. With a bit of luck, some clever choices of radical expressions might allow you to make that distinctions at the level of radicals.
For concrete numbers, all of this should be feasible to do, and yield reasonable expressions if you can accept some rounding. For purely symbolic calculations, the expressions involved would likely be too large to be of much use to humans. Which is why I didn't print them here. There is always the chance someone can think of a clever way to express them more concisely. Computers of course are less easily daunted by a single formula filling a screen page, so if you want to construct a symbolic expression to feed to a computer, that is also possible.
By only putting $\cos\theta$ and $\sin\theta$ into my formula, the derivation from a Cartesian representation shows. It's possible that by using some other functions of the angle you can come up with a simpler representation. But anything you do has to be complex enough to still describe an eighth degree algebraic curve, so I wouldn't expect anything radically simpler.