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.
Best Answer
They're not. No two different circles centered at the origin contain any of the same points. There are uncountably many circles (specifically, one for each real number, corresponding to the radius) , so most circles contain no rational points at all.
We can find some more specific examples. Specifically, any rational point $(a, b)$ on a circle of radius $r$ centered at the origin satisfies $a^2+b^2=r^2$. In particular, $r^2$ must be rational. There are also radii whose squares are rational where there are no rational points. Clearing denominators, say multiplying by some $c^2$ to do so, we have that $c^2r^2$ is a sum of two squares. If $r^2$ is an integer, then $r^2$ must be a sum of two squares, since an integer is a sum of two squares if and only if its prime factorization doesn't contain an odd power of a prime congruent to $3$ mod $4$. $r^2$ was arbitrary, so if we choose it not to be a sum of two squares we get circles with no rational points.