The question that I am referring to is:
Let $D$ be the closed disc bounded by the circle with equation
$$(x-6)^{2} + y^{2} = 36$$
I am asked to find the parameterization of the boundary.
I am a bit confused on how to do this. I know that $x^2 + y^2 = 36$ which means $r = 6$.
Since $x = r\cos(\theta)$ and $y = r\sin(\theta)$, how would the $(x-6)^2$ be parameterized?
HINT
In general, if we are given the circumference \begin{align*} (x-a)^{2} + (y-b)^{2} = r^{2} \end{align*}
we can parametrize it as follows
\begin{align*} \begin{cases} x = a + r\cos(\theta)\\\\ y = b + r\sin(\theta) \end{cases} \end{align*}
where $\theta\in[0,2\pi]$. Can you apply it to the example proposed?