<s>[INST] 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 $(0.-3)$?
 [/INST] Let's start with the value of $\theta$, in radians, as a function of time. Because the particle is travelling at a constant velocity, revolving in $9$ seconds, $\dfrac{d\theta}{dt} = -\dfrac{2\pi}{9}$. The minus sign is because the particle is travelling clockwise. Then 
$\theta(t) = -\dfrac{2\pi}{9}t + \theta_0$. Since the angle of the point $(-3,0)$ is $-\pi$, we must have 
$-\pi = \theta(0) = \theta_0$
So
$\theta(t) = -\dfrac{2\pi}{9}t - \pi$.
So 
$(x,y) = \left(3 \cos\left( -\dfrac{2\pi}{9}t - \pi\right),
               3 \sin\left( -\dfrac{2\pi}{9}t - \pi\right)
         \right)
       = \left( -3\cos\left( \dfrac{2\pi}{9}t\right),
                 3\sin\left( \dfrac{2\pi}{9}t\right)
         \right)$
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