Given is a $2\times n$ matrix $A$ with orthogonal rows and $c>0$ ($n$ is actually $4$ in my case, however, I don't think that matters):
Question: What is the pre-image of the linear operator $A: \mathbb{R}^n \to \mathbb{R}^2$ of a circle of radius $c$ in $\mathbb{R}^2$, i.e., what can be said about the set $$ Y = \bigl\{ x \in \mathbb{R}^n \mathrel{:} Ax = (y_1, y_2)^\top, \; y_1^2 + y_2^2 = c^2 \bigr\}? $$
In analogy to the result that the image of a circle for a linear, orthogonal mapping from $\mathbb{R}^2$ to $\mathbb{R}^2$ is an ellipse, I would conjecture that the set $Y$ above is some sort of a cylinder in $\mathbb{R}^n$ with an ellipse as a base. Clearly, this is the case if $A$ is a projection from $\mathbb{R}^n$ to $\mathbb{R}^2$.
Let the two row vectors be $u_1, u_2 \in R^n$. Writing $\cdot$ for the standard dot product, the condition for a vector $x$ to belong to the preimage is (u_1 \cdot x)^2 + (u_2 \cdot x)^2 = c^2.
Now select a system of rectangular coordinates $x = (x_1, \dots, x_n)$ in which $u_1 = (\lambda_1, 0, \dots, 0)$ and $u_2 = (0, \lambda_2, 0, \dots, 0)$, where $\lambda_1, \lambda_2 \geq 0$.
Then in the new coordinate system, the equation satisfied by $x$ can be written as
$$\lambda_1^2 x_1^2 + \lambda_2^2 x_2^2 = c^2,$$
which is obviously an elliptic cylinder in general. When $|u_1| = |u_2| > 0$, we have a right circular cylinder. When exactly one of $u_1, u_2$ is zero, we have two parallel hyperplanes, and when they are both zero the preimage is empty.