Question regarding parametric equations A circle of radius 2 rolls along a flat surface at a constant rate, as shown below. A point which has a distance of 3 from the center of the circle is fixed with respect to the circle.

Let $A$ be the midpoint of the arc, and let $B$ be a point where the path of the point intersects the surface. Let $v_A$ and $v_B$ be the speeds of the point at $A$ and $B,$ respectively. Compute $\frac{v_A}{v_B}.$
How do I set up a parameterization and then find the speeds?
 A: The center of the circle follows this path.
$x = 2\theta, y = 2$
If the radius was something other than 2, you would plug that value.
A point moving in a circular path of radius 3.
$x = 3\cos \theta, y = 3\sin \theta$
But this is a counter-clockwise rotation, and first set of equations reflect a clockwise rotation.  And lets change these such that the minimum y values is associated with $t = 0$
$x = -3\sin \theta, y = -3\cos \theta$
And we can just add these two components of motion together.
$x = 2\theta - 3\sin \theta, y = 2 - 3\cos\theta$
To find velocity.
$v_x = \frac {dx}{d\theta} = 2 - 3\cos\theta\\
v_y = \frac {dy}{d\theta} = 3\sin\theta$ 
When $\theta = 0, (v_x,v_y) = (-1,0)$
speed $\|v\| = 1$
To find $A$ we will need to find a second value of $\theta$ where
$x = 2\theta_2 - 3\sin \theta_2 = 0$
You will need numerical methods.
Once you have $\theta_2$
finding $v(\theta_2)$ is fairly simple.
A: The description of the problem is ambiguous (what, for instance, is the arc of which $\ A\ $ is the mid-point), and my interpretation seems to be different from those adopted by the other two answers.  I'm assuming the point $\ A\ $ is at one of the points where the arc is at its maximum height of $\ 5\ $ units above the surface on which the circle is rolling, as depicted in the diagram below.  I'll take the $\ x$-axis to be along this surface in the same direction as the circle is rolling, the origin $\ O\ $ to be the point on the surface immediately below $\ A\ $, and the $\ y$-axis to be along the line from $\ O\ $ to $\ A\ $.  After the circle has rotated through an angle of $\ \theta\ $ radians, its centre will have moved $ \ 2\theta \ $ units, and the coordinates of the moving point will be
\begin{align}
x&=2\theta+3\sin\theta\\
y&=2+3\cos\theta \ ,
\end{align}
as illustrated in the diagram below.  The components of its velocity will be
\begin{align}
\dot{x}&=\dot{\theta}\,(2+3\cos\theta)\\
\dot{y}&=-3\dot{\theta}\sin\theta \ ,
\end{align}and its speed will be
$$
v=\sqrt{\dot{x}^2+\dot{y}^2}=\dot{\theta}\sqrt{\mathstrut4+12\cos\theta+9}\ .
$$
At $\ A\ $ we have $\ \theta=0\ $, so $\ v_A=5\dot{\theta}(0)\ $, and at any point $\ B\ $ where the moving point is crossing the surface, we have $\ y=0\ $, or $\ \theta=-\frac{2}{3}\ $, so
$\ v_B=\sqrt{5}\dot{\theta}\big(t_B\big)\ $. If the motion is uniform, then $\ \dot{\theta}\ $ is constant, and
$$
\frac{v_A}{v_B}=\sqrt{5}\ .
$$

