Particle starts at $(0,-3)$ and moves clockwise around origin on graph $x^2+y^2=9$, find parametric equation Question

particle starts at $(0,-3)$ and moves clockwise around origin on graph $x^2+y^2=9$, revolve in $9$ seconds find parametric equation in term of $t$.

What I've done so far:
I first thought that the graph ought to be $x^2+y^2=9$
so then I say that $x=\frac{9}{2}\cos{t}=x$ and $y=\frac{9}{2}\sin{t}$
but then this particle in this graph travels CCW so then I change it to : 
$x=\frac{9}{2}\cos{t}=x$ and $y=\frac{9}{2}\sin{-t}$
but then I found out that when I plug in $t=0$, I do not get -3
How do I phase shift this parametric equation so that it satisfies the fact that the particle starts at $(0.-3)$?

Let's start with the standard clockwise parametrization,
$$x=3\cos (t)$$
$$y=-3\sin (t)$$
With $t \in [0,2\pi]$. This starts at $(3,0)$ and then moves clockwise. 
We want it to start at $(0,-3)$. If we rotate $(3,0)$ clockwise $90$ degrees we get $(0,-3)$. So shift $\frac{\pi}{2}$ radians clockwise. Let,
$$x=3\cos (t+\frac{\pi}{2})$$
$$y=-3\sin (t+\frac{\pi}{2})$$
Now we need to deal with the fact that it should take $9$ seconds to do a full revolution. Right now it takes $2\pi$ seconds because the period of both is $2\pi$. If the new period is $9$ then $2\pi$ over the horizontal shift $a$ should be $9$.
$$\frac{2\pi}{a}=9$$
$$a=\frac{2\pi}{9}$$
So,
$$x=3\cos (\frac{2\pi}{9}t+\frac{\pi}{2})$$
$$y=-3\sin (\frac{2\pi}{9}t+\frac{\pi}{2})$$
