Maximum speed in a circular orbit? Visualize two points:  
$O\equiv(0\mid 0)$ and $D\equiv(d\mid 0)$.  
The two are $d$ units apart.  
Visualize a movable rod whose endpoints, $C_O$ and $S_O$, are a unit apart. $C_O$ always coïncides with $O$ and $S_O$ orbits $O$.
  
Visualize a movable rod whose endpoints, $C_D$ and $S_D$, are a unit apart. $C_D$ always coïncides with $D$ and $S_D$ orbits $D$.  
  
Visualize a movable rod whose endpoints, $R_O$ and $R_D$, are $d$ units apart. $R_O$ coïncides with $S_O$, and $R_D$ coïncides with $S_D$.  
  
I believe that such an arrangement is called a linkage. For our purpose $S_O$ (and $R_O$) can be considered as situated at $(\cos\theta\mid \sin\theta)$. $\theta$ is the parameter whose value determines the locations of all moving points in this problem.  
  
There are two such linkages.  
  
In the first, the rod connecting $S_O$ to $S_D$ is always parallel to the line connecting $O$ to $D$. In this case, $S_D\equiv(d+\cos\theta\mid \sin\theta)$. Not interesting, therefore ignorable.  
  
In the second, the rod connecting $S_O$ to $S_D$ is never parallel to the line connecting $O$ to $D$. In this case, $S_D\equiv\left({{(d^2-1)(d-\cos\theta)}\over{d^2+1-2d\cos\theta}}\mid{{(d^2-1)(-\sin\theta)}\over{d^2+1-2d\cos\theta}}\right)$.
If one considers the direction of rotation of $OS_O$ to be positive $+$, then the following is true:
If $d\gt 1$, then the direction of rotation of $DS_D$ is $+$.
If $d\lt 1$, then the direction of rotation of $DS_D$ is $-$.
If $d=1$, then $S_D\equiv (0\mid 0)$ — except when $\theta\equiv 0\mod 2\pi$, when $S_D\equiv{\left({0\over 0}\mid{0\over 0}\right)}$.  
QUESTION: For a given positive value of $d$, what value of $\theta$ imparts the maximum angular velocity, $+$ or $-$, to $DS_D$?  
I am indebted to Dr. Jyrki Lahtonen for his helpfulness!!!!
[Editor's note: The picture I have in mind, JL.] 
The animations below have the points $S_O$ and $S_D$ marked with black dots. As the distances from the origin $O$ to $S_O$, from $S_O$ to $S_D$ and from $S_D$ to $D$ are all fixed, we can imagine them being connected by rigid rods. I took the liberty of adding those rods to the pic. 
In the first animation, $d={2\over 3}$. 

And in the second, $d={4\over 3}$ (the size of the animation is scaled down by a factor of two in comparison to the one above).

[Editor's note: I put the motivational speech below (from the OP) back in here, because it may be needed to make the question meet the site standards, JL]
ADDENDUM! It was requested that I clarify my question, so I shall try to eliminate abstractions in favor of concrete objects …  1. Replace the Cartesian plane with a sheet of marine plywood.  2. On the plywood, draw a line segment $d$ units long. Label its endpoints $O$ and $D$. Drill a hole at each endpoint. Fit a dowel snugly into each.  3. Take three paint mixing paddles. On two of them, draw lines $r$ units long. On one of these, label the endpoints $O$ and $S_O$; on the other, label the endpoints $D$ and $S_D$. Drill holes at all four of these endpoints. On the remaining paint mixing paddle, draw a line $d$ units long. Label its endpoints $S_O$ and $S_D$. Drill a hole at each. 4. Fit the hole $O$ onto the dowel $O$. Fit the hole $D$ onto the dowel $D$. Fit dowels into $S_O$ and $S_D$, but not completely through. 5. Fit the $S_O$ hole onto the $S_O$ dowel. Fit the $S_D$ hole onto the $S_D$ dowel. 6. Saw off any excess dowel that would interfere with the movement of the assembly. (Yes, it is supposed to move.) 7. I envision these rods as being made of untreated white pine.
NOTA BENE: For simplicity's sake, I have gotten rid of the variable $r$ and replaced it with unity.
 A: I'm breaking my promise not to answer this as no one else bites. This is largely just a sketch of an argument explaining why the maximum is obtained at $\theta=0$, when $S_O$ is as close to $D$ as possible, moving tangentially. A calculus solution by differentiation is possible, but I don't want to do that.
Here's a variant of the animation. This time $d\approx 1.54$. I added the linkage discarded by the OP in red, because it aids understanding what is going on. Furthermore I added a green `rubber band' connecting the points $D$ and $S_O$.

As expected, the `uninteresting' red dot, call it $S_D'$, revolves about the pivot $D$ with exactly the same angular rate as the point $S_O$ about the pivot $O$. This is a consequence of the fact that $DS_D'S_OO$ is always a parallelogram. But let's look at the picture more closely. We see that the black dot $S_D$ is always the mirror image of $S_D'$ with respect to the green line. The mirror symmetry is a geometrically obvious consequence of the fact that the points $S_D$ and $S_D'$ are the two points on the plane with the prescribed distances to $D$ and $S_O$.
Therefore the angular velocity of the green line (about $D$) is the average of the angular velocities of $S_D$ and $S_D'$ (the green line bisects the angle $\angle S_DDS_D'$).
So
$$
\omega_{S_D}=2\omega_{green}-\omega_{S_D'}.
$$
As $\omega_{S_D'}$ is constant this means that $\omega_{S_D}$ is maximized simultaneously with $\omega_{green}$. But $\omega_{green}$ is the angular velocity of $S_O$ about $D$.  We can apply the principle: angular velocity = tangential component of the orbital speed vector divided by the distance. As the orbital speed of $S_O$ is constant it happens that the tangential component is maximized at the point of the closest approach, so the numerator reaches its maximum and the denominator reaches its minimum simultaneously at $\theta=0$. Therefore the ratio reaches its maximum at the closest approach of $S_O$ to $D$.
