Trying to figure out formula to find the angle to rotate a pivot so that an attached point with a direction is '"facing" a third point Sorry for the title, I am finding it hard to describe the problem I am trying to solve...
I am trying to figure out the angle I need to rotate a point in 2d space which is currently "facing" another point in the distance so that another point which is offset from the pivot position can now face the point in the distance. Hopefully the below image explains this better than words:

I am trying to figure out angle (?) which the pivot (a) has to rotate by so that point (c) is now facing point (b).
 A: You're almost there. Here's the right-hand side of your diagram with a few added markings:

The labels $a$ and $b$ are restored to the pivot position and the point in the distance;
the label $x$ is restored to the segment whose length represents the offset distance, and the label $z$ is restored so that it indicates the distance from $a$ to $b$.
Note that the line $ab$ crosses two parallel lines and therefore makes the same angle $B$ with each line.
The angle $B$ at the distant point $b$ is one angle of a right triangle whose hypotenuse is $z$ and whose opposite leg is $x$;
therefore
$$ \sin B = \frac xz. $$
This can be inverted by using the arc sine function (the inverse of the sine function):
$$ B = \arcsin\left(\frac xz\right) . $$
Now as we have already noted, both angles labeled $B$ have the same angular measure;
the angle $B$ between the lines meeting at $a$ is the angle you want.
Note that the distance $y$ in the original figure is irrelevant as long as $y^2 < z^2 - x^2.$
If $y^2 = z^2 - x^2$ then the circle passes through $b$ and if
$y^2 > z^2 - x^2$ then $b$ is inside the circle, so it is not possible in those cases for $c$ to "face" $b$ as shown in the figure.
