Vector equation of a circle

Question

Recently I came across the vector equation of a circle as $$r(t)=a+b \cos(t) +c \sin(t)$$ I want to prove it but I am unable to proceed . The only thing I feel is that $a$ which is actually $(a_x,a_y,a_z)$ is the center. I am not sure if I am correct. Please help me with the proof.

Answer 1

We can get rid of the first term $a$ which adds only a "final" translation.

The result is that $r=b \cos(t) +c \sin(t)$ where $b,c$ are 3D vectors is a parametric equation of an ellipse situated evidently in the vector plane (P) defined by $b$ and $c$.

Why is it an ellipse ?

Let us consider the 2D problem in plane (P) with 2D notations. It means that vectors $B$ and $C$ being given in this 2D plane, and we are looking for the nature of the curve described by

$$R=B \cos(t) + C \sin(t)\tag{2}$$

with coordinates et $R=(x,y)$, $B=(B_x,B_y)$ and $C=(C_x,C_y)$

(2) is equivalent to the following system of equations :

$$\begin{cases}x&=&B_x \cos(t)+C_x\sin(t)\\y&=&B_y \cos(t)+C_y\sin(t)\end{cases} \ \ \iff \ \ \begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}B_x&C_x\\B_y&C_y\end{pmatrix}\underbrace{\begin{pmatrix}\cos(t)\\ \sin(t)\end{pmatrix}}_{\text{unit circle}}\tag{3}$$

therefore an ellipse as the affine image of the unit circle.

Remark : we will get a circle is the above $2 \times 2$ matrix is the matrix of a similitude, hopefully composed with a symmetry, which means that the columns of the matrix should be orthogonal with the same length.

Coming back to 3D, the last sentence means that we get a circle iff $b \perp c$ and $\|b\|=\|c\|$.