Geometric transformation on circle equation

Question

Suppose that I have variables $x_1,x_2$ and following circle equation: $x_1^2+x_2^2=1$. Now I have a matrix $A$ which will be used to transform my circle equation. $Z=AX$ where $X$ is a vector with $x_1,x_2$ and $A$ is the transformation matrix. What will the final circle look like and how can I compute its equation? (The final shape can be an ellipsoid).

Answer 1

You can parametrize your circle as:

$$x_1 = \cos(t)$$ $$x_2 = \sin(t)$$

for $0\leq t < 2\pi$. The column vector for a point at a given value of $t$ is: $$ \mathbf{x}(t) = \left(\begin{array}{c} \cos(t) \\ \sin(t) \end{array}\right) $$ again for any $0\leq t < 2\pi$.

For matrix $A = [ a_{ij} ] $, the image of all points $\mathbf{x}(t)$ can be written:

$$ A\mathbf{x}(t) = \left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right)\left(\begin{array}{c} \cos(t) \\ \sin(t) \end{array}\right) $$

The resulting set of points can be parametrized as $x_1 = a_{11}\cos(t) + a_{12}\sin(t)$ and $x_2 = a_{21} \cos(t) + a_{22} \sin{t}$. This is an explicit equation for the image of the circle under $A$.

However, it doesn't really illustrate what happens geometrically. What kind of things can a matrix $A$ do to a shape in the plane? It can do any combination of rotating, scaling, shearing, reflection, and projection. Maybe you can convince yourself that applying any of the first four to a circle yields an ellipse. However, if the matrix represents a projection transformation, the image of the circle will be a line segment. I'm sure you can demonstrate all of these things algebraically, but it might be more rewarding to graph a few examples.