Is it fair to say that Kepler's equation involves squaring the circle? I'm trying to understand the degree to which I'm at sea, anchored, or on firm ground, and how to firm up my understanding as needed.
I think Kepler's equation reduces to a task (call it $t$, which I'll describe better shortly) that is equivalent to finding a circle with equal area as a square. So:
1) Is it appropriate and correct to say that solving Kepler's equation reduces to squaring the circle?
I suppose you could tell that I'm not a real geometer, or algebraist, or:
2) What area of mathematics am I grasping at here? And am I using appropriate language?
3) Is all of this a reasonable way to explain why Kepler's equation is said to be transcendental? Is it also a reasonable way to explain why "how to square the circle" is an interesting question even now that we know there's no method using the compass and unruled straightedge (or, what I read is equivalent, the "quadratic closure of the rationals")?
Where I mentioned a task $t$, what I have in mind is finding a region partly bounded by a circular arc with equal area as a triangle. I got this notion when I read "Computing the position as a function of time". It presents a figure that it calls a geometric construction:

So when I ask in (1) whether I'm correct, I guess what I actually want to be correct about is that point $x$ (or equivalently $P$ or $d$) in the so-called construction is not "constructible" from points $c$ the center of the ellipse, point $S$ a focus of the ellipse, and point $y$ on the circle around $c$ of radius as large as the semi-major axis of the ellipse and with angle $Scy$ proportional to the time since periapsis, that is, the nearest approach of $P$ to $S$.
It is claimed that Kepler's equation provides that the area of region $zSx$, bounded by two line segments and arc $zcx$, equals the area of circular sector $zcy$. Removing the region in common and giving name $k$ to the point where $cy$ intersects $Sx$, we have equal areas of triangle $Sck$ and region $ykx$ that is bounded by two line segments and arc $ycx$. Then task $t$ is to find $x$ such that we get the equal areas that we have just mentioned where $k$ is determined as we have just defined.
As to squaring the circle, well, it is trivial to find a square with the same area as triangle $Sck$ and not difficult, I imagine, to find a circle with the same area as the region $ykx$.
 A: After a brief chat with @OrangeHarvester I think it's this way: I was overreaching for a connection between Kepler's laws of planetary motion and the non-constructible.
Yes, Kepler's equation is in the nature of squaring the circle, in that it is transcendental, due to the use of the sine function.
I just learned, as I did not previously know, that the constructible numbers are a proper subset of the algebraic numbers, which are the complement of the transcendental numbers. The "impossibility" of squaring the circle, we remember, is that it requires using non-constructible numbers.
But no, Kepler's equation is not so special in that way. There were transcendental equations long before Kepler.
However, if we prefer, then Kepler's equation is a particularly lovely example of a transcendental equation, in the sense that de gustibus non est disputandum: in matters of taste there is no proof.
A: “Morbus cyclometricus” may eventually translate to “sanitas cyclometricus” if this new perspective of Pi confirms the value of all human effort to “square the circle”: http://www.aitnaru.org/images/Pi_Corral.pdf (file attached to web page)
The scalene triangle does exist in the range of possible squares of the circle from smallest to largest. Does geometry exist to define that unique triangle without trial and error and according to the Greek rules for this challenge?
This unique triangle, inscribed within a circle, should at least help promote a new generation of ideas for food and festivities for Pi Day 2014 … and new ideas for squaring the circle! 
