Translate a point on a circumference If I have a point $A$ on the circumference of a circle with origin $O$ and radius $r$, how would I find the coordinates of point $B$, which is also on that circumference, but is $d$ units away from point $A$ in a clockwise direction, that is, $A$ is translated $d$ units along the circumference.
 A: All we have to do is apply the clockwise rotation matrix with the appropriate angle $\theta$ to the coordinates of $A$. 
First we need to figure the angle  to rotate by - that will be $\theta=\dfrac{d}{r}$ (this is the appropriate fraction of the circumference in radians).
Now we need to actually perform the rotation. Let the coordinates of $A$ be $(a_1,a_2)$ and the coordinates of the center $C$ be $(c_1,c_2)$ (I am assuming that by $O$ in your initial question, you meant some arbitrary center coordinate). 
We are going to treat the center as the origin of the plane and rotate around that to get $B$. Since we are now treating $C$ as the origin, we only care about where $A$ lies relative to $C$ - that is at coordinates $(a_1-c_1,a_2-c_2)$. Finally we are ready to rotate using the clockwise rotation matrix!
Here we go:
$$\left[\begin{matrix}\cos\theta&\sin\theta
\\-\sin\theta&\cos\theta
\end{matrix}\right]\left[\begin{matrix}a_1-c_1\\a_2-c_2\end{matrix}\right]=\left[\begin{matrix}(a_1-c_1)\cos\theta+(a_2-c_2)\sin\theta\\-(a_1-c_1)\sin\theta+(a_2-c_2)\cos\theta\end{matrix}\right]$$
Finally we add back the coordinates of the center so we end up in our original coordinate system. This gives the final coordinates of $B$ as
$$(b_1,b_2)=((a_1-c_1)\cos\theta+(a_2-c_2)\sin\theta+c_1,-(a_1-c_1)\sin\theta+(a_2-c_2)\cos\theta+c_2)$$
And don't forget to use radians!
A: By performing a suitable rotation of axes, we can always assume point $B$ is at $(r, 0)$.
Let $B’(h, k)$ be the target after the said transformation. The transformation rotates (in this case we also assume that it is in anti-clockwise direction, but the logic equally applies even if it is in the clockwise direction) a central angle $\theta$ (in radian measure) subtending the arc of length $d$.
Then, $(h, k) = (r \cos \theta, r \sin \theta)$, where $\theta$ is given by
$$
\theta = \frac {d}{r}
$$
A: You have not given position of A fully. Let A have polar coordinates $ (r, \alpha)$. You rotate A by $ \theta = \dfrac {d}{r}$ So new coordinates are
$$ (r, \alpha \pm \theta) $$ 
depending on clockwise or anti-clockwise rotation sense; you can convert to Cartesian if you wish.
