Line
You already have a solution for that.
Circle
A circle is described by an equation like this:
$$
0 = (x-c_x)^2 + (y-c_y)^2 - r
= x^2 + y^2 + (-2c_x)x + (-2c_y)y + (c_x^2+c_y^2-r)
$$
Multiplying that equation by a real factor and renaming the coefficients, you obtain an equation of the following form:
$$\alpha(x^2+y^2) + \beta x + \gamma y + \delta=0$$
Three points in the plane define a circle. They define the parameters $\alpha$ through $\delta$ up to a common scale factor. Adding a fourth point, you have two options: Either that point lies on the circle, then it fits the same 1d space of parameters. Or it does not lie on the same circle, then the only way to satisfy all four equations is by choosing all four parameters equal to zero.
So you can check cocircularity by evaluating the determinant of this system of equations:
$$\begin{vmatrix}
x_1^2+y_1^2 & x_1 & y_1 & 1 \\
x_2^2+y_2^2 & x_2 & y_2 & 1 \\
x_3^2+y_3^2 & x_3 & y_3 & 1 \\
x_4^2+y_4^2 & x_4 & y_4 & 1
\end{vmatrix}=0$$
If you write this out, you get something pretty bulky:
\begin{align*}
x_{2}^{2} x_{3} y_{1} - x_{2} x_{3}^{2} y_{1} - x_{2}^{2} x_{4} y_{1} +
x_{3}^{2} x_{4} y_{1} + x_{2} x_{4}^{2} y_{1} - x_{3} x_{4}^{2} y_{1} -
x_{1}^{2} x_{3} y_{2} + x_{1} x_{3}^{2} y_{2} \\{}+ x_{1}^{2} x_{4} y_{2} -
x_{3}^{2} x_{4} y_{2} - x_{1} x_{4}^{2} y_{2} + x_{3} x_{4}^{2} y_{2} -
x_{3} y_{1}^{2} y_{2} + x_{4} y_{1}^{2} y_{2} + x_{3} y_{1} y_{2}^{2} -
x_{4} y_{1} y_{2}^{2} \\{}+ x_{1}^{2} x_{2} y_{3} - x_{1} x_{2}^{2} y_{3} -
x_{1}^{2} x_{4} y_{3} + x_{2}^{2} x_{4} y_{3} + x_{1} x_{4}^{2} y_{3} -
x_{2} x_{4}^{2} y_{3} + x_{2} y_{1}^{2} y_{3} - x_{4} y_{1}^{2} y_{3} \\{}-
x_{1} y_{2}^{2} y_{3} + x_{4} y_{2}^{2} y_{3} - x_{2} y_{1} y_{3}^{2} +
x_{4} y_{1} y_{3}^{2} + x_{1} y_{2} y_{3}^{2} - x_{4} y_{2} y_{3}^{2} -
x_{1}^{2} x_{2} y_{4} + x_{1} x_{2}^{2} y_{4} \\{}+ x_{1}^{2} x_{3} y_{4} -
x_{2}^{2} x_{3} y_{4} - x_{1} x_{3}^{2} y_{4} + x_{2} x_{3}^{2} y_{4} -
x_{2} y_{1}^{2} y_{4} + x_{3} y_{1}^{2} y_{4} + x_{1} y_{2}^{2} y_{4} -
x_{3} y_{2}^{2} y_{4} \\{}- x_{1} y_{3}^{2} y_{4} + x_{2} y_{3}^{2} y_{4} +
x_{2} y_{1} y_{4}^{2} - x_{3} y_{1} y_{4}^{2} - x_{1} y_{2} y_{4}^{2} +
x_{3} y_{2} y_{4}^{2} + x_{1} y_{3} y_{4}^{2} - x_{2} y_{3} y_{4}^{2} &=0
\end{align*}
Note that four points on a line will also satisfy this equation, and the solution will have $\alpha=0$. If you don't want to include this case, check for lines first.
Rectangle
Two vectors are orthogonal if their dot product is zero. Do this for three pairs of edge vectors and you have checked three corner angles. If they are all right angles, then so is the fourth and you have the corners of a rectangle.
For points on the edge of a rectangle: you will always find a rectangle which has the four points as elements of its edges as long as the points are in convex position. Start by connecting two points with a line, then extend that to the smallest bounding rectangle of the orientation indicated by that line. There are examples where the first choice of line would lead to one of the other points inside the resulting rectangle, but as far as my mental experiments go (although this is no proof), you can always find a different starting edge such that things work out.
Triangle
For four points to be the corners of a triangle, two of them have to be the same.
You will always find a triangle which has the four points on its edges as long as the points are in convex position. Simply take the convex hull, which is a quadrilateral in this case, and extend two of its edges to form a triangle.
The key observation is that the circumcircle of the polygon with $n$ sides is the incircle of the polygon with $n+1$ sides; thus $$R_n = r_{n+1}.$$ But we also have the relationship $$\cos \frac{\pi}{n} = \frac{r_n}{R_n}$$ in a given regular $n$-gon, thus we have $$\frac{R_N}{r_3} = \prod_{n=3}^N \frac{R_n}{r_n}$$ and as $N \to \infty$, $$\frac{R_\infty }{r_3} = \prod_{n=3}^\infty \sec \frac{\pi}{n} \approx 8.7.$$ This suggests that were the process of circumscribing regular polygons and circles were continued indefinitely, the figure reaches a limiting, finite size. If we were to perform the reverse and continually inscribe polygons, then this too would result in a limiting incircle of positive radius, which I will leave to others to calculate.
Here is a reference from Mathworld: http://mathworld.wolfram.com/PolygonCircumscribing.html
Best Answer
I can suggest some names, which I have learned from others who research these things.
There are 4 types of ring torus objects in 4D, which can be seen visualized here. The general name of hypertorus works well:
Spheritorus : sphere-bundle over the circle :$S^2$ x $S^1$ $$\left(\sqrt{x^2+y^2} -a\right)^2 +z^2+w^2 = b^2$$
Torisphere : circle-bundle over the sphere : $S^1$ x $S^2$ $$\left(\sqrt{x^2+y^2+z^2} -a\right)^2 +w^2 = b^2$$
3-torus : circle over circle over circle : $T^3$ $$\left(\sqrt{\left(\sqrt{x^2+y^2}-a\right)^2+z^2}-b\right)^2+w^2 = c^2$$
Tiger : circle-bundle over the flat 2-torus (Clifford torus) $$\left(\sqrt{x^2+y^2} -a\right)^2 +\left(\sqrt{z^2+w^2} -b\right)^2 = c^2$$
As for the bar -> cylinder -> duocylinder, I'm not sure exactly what sequence you are using here. The best fit I can see is describing a specific bisecting rotation around an n-1 plane into n+1 dimensions. In this case, the next 5D shape is called a Cylspherinder , a cartesian product of a $D^2$ and $D^3$ (solid disk times solid sphere). But, you can also make a Spherinder (sphere prism, another type of 4D cylinder) from a rotation of a cylinder into 4D.
Cylinder: $\left|\sqrt{x^2+y^2} -z\right|+\left|\sqrt{x^2+y^2} +z\right| = a$
Duocylinder: $\left|\sqrt{x^2+y^2} -\sqrt{z^2+w^2}\right|+\left|\sqrt{x^2+y^2} +\sqrt{z^2+w^2}\right| = a$
Spherinder : $\left|\sqrt{x^2+y^2+z^2} -w\right|+\left|\sqrt{x^2+y^2+z^2} +w\right| = a$
Cylspherinder : $\left|\sqrt{x^2+y^2+z^2} -\sqrt{w^2+v^2}\right|+\left|\sqrt{x^2+y^2+z^2} +\sqrt{w^2+v^2}\right| = a$
I guess you can call these n-cylinders, but there are even more types of these than just product of n-balls and n-cubes. You can also include the product of n-ball and n-simplex as well. In fact, any shape with both flat and curved cells can fit into this group (product of 2-ball (and higher) with any genus-0 object)
Cyltrianglinder : $\left|\big||x|+2y\big|+|x| -2\sqrt{z^2+w^2}\right|+\left|\big||x|+2y\big|+|x| +2\sqrt{z^2+w^2}\right| = a$