Somewhat more elegant would be:

plt = ContourPlot[{Abs[x + I y] == 2,    Abs[x + (x + I y)^2 + I y] == 2}, {x, -2, 2}, {y, -2, 2}]
curve = 2;
ind = Cases[List @@ plt, Line[x_] -> x, Infinity][[curve]];
ppp = List @@ Cases[List @@ plt, GraphicsComplex[x_, y__] -> x][[1]];
points = ppp[[#]] & /@ ind;
plt2 = ListPlot[points]
Show[plt2, plt]

You can get the implicit relation as below.

ee = x + (x + I y)^2 + I y;
eesq = ComplexExpand[Expand[ee*Conjugate[ee]]]

(* Out[170]= x^2 + 2 x^3 + x^4 + y^2 + 2 x y^2 + 2 x^2 y^2 + y^4 *)

A parametric form is a different matter though. Not sure what might work for that.

Yes, it relates to Mathematica. Apologies for not being more specific. I am trying to use Mathematica to answer the questions below. We generate the first two iterations of the Mandelbrot Set with:

ContourPlot[{ Abs[x + I y] == 2,  Abs[x + (x + I y)^2 + I y] == 2}, {x, -2, 2}, {y, -2, 2}]

The first iteration is Abs(x + I y) = 2 . It corresponds to a simple circle, { x(t)= 2 Cos(t) , y(t)= 2 Sin(t) } . Finding a parameterization { x(t) , y(t) } that corresponds to the 2nd iteration, Abs(x + (x + I y)^2 + I y) = 2 proves more difficult. (it's not an ellipse, though it looks like one).

Question 1: Right-clicking the plot allows you to Get Coordinates for one specific point. Is there a way to export a list of many coordinates for the curve? Quadruple clicking on the curve highlights many point on the curve.

How would we export the coordinates of those points to a list?

Question 2: Using Reduce can help us solve for x or y, for instance: Yet somehow, y = (all that) is equivalent to 2 Sin(t) when mapping from {real, imaginary} to {x,y}. Does anyone know of a way to show that this is true ? I would think it's straightforward, since it's just a simple circle. But i'm lost. Thank you.