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,

$$x=3\cos (\frac{2\pi}{9}t+\frac{\pi}{2})$$

$$y=-3\sin (\frac{2\pi}{9}t+\frac{\pi}{2})$$
$$
x=-3\sin\left(\dfrac{2\pi}{9}t\right), \quad y=-3\cos\left(\dfrac{2\pi}{9}t\right)
$$