Calculating point around circumference of circle given distance travelled It is the end of the day and my brain just can't cope anymore. Can anyone help with, what I hope, is a simple question? 
Given a point on a circle ($x$, $y$ coordinates) how can I calculate the coordinates of another point at a given distance around the circumference of the circle? Known variables are the starting point, distance traveled and radius of the circle.
 A: Let the center of the circle be $(a,b)$.  Drag the center to the origin. This procedure drags our given point $(x,y)$ to $(x-a,y-b)$.
For simplicity of notation, let $u=x-a$, $v=y-b$.  Now we determine the angle that the positive $x$-axis has to be rotated through (counterclockwise) to hit the line from the origin to $(u,v)$.  Call this angle $\theta$.
Then $\theta$ is the angle, say in the interval $(-\pi,\pi]$, whose cosine is $u/r$, and whose sine is $v/r$, where $r$ is the radius of the circle.  (This was already known; it also happens to be $\sqrt{u^2+v^2}$.)  So from now on we can take $\theta$ as known.  But we have to be careful to take the signs of $u$ and $v$ into account when calculating $\theta$. 
Let $D$ be the distance travelled. Assume that we are travelling counterclockwise. Then the angle of travel is (in radians) equal to $D/r$. Let $\phi$ be this angle. If we are travelling clockwise, just replace $\phi$ by $-\phi$.  So from now on we can take $\phi$ as known.  
After the travel, our angle is $\theta+\phi$.  This means that we are at the point with coordinates
$$(r\cos(\theta+\phi), \: \: r\sin(\theta+\phi)).$$
Now transform back, by adding $(a,b)$ to the point.  We obtain
$$(a+r\cos(\theta+\phi),\:\: b+r\sin(\theta+\phi)).$$
All the components of this formula are known, so we can compute the answer.
Comment:  Note that 
$$\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi.$$
But we know that $r\cos\theta=u$ and $r\sin\theta=v$.
Thus 
$$a+r\cos(\theta+\phi)=a+u\cos\phi-v\sin\phi.$$
Similarly, 
$$\sin(\theta+\phi)=\cos\theta\sin\phi+\sin\theta\cos\phi,$$
and therefore
$$b+r\sin(\theta+\phi)=u\sin\phi+v\cos\phi.$$
Thus an alternate (and for many purposes simpler) version of the answer is
$$(a+(x-a)\cos\phi-(y-b)\sin\phi,\:\: b+(x-a)\sin\phi+(y-b)\cos\phi),$$
where $\phi=D/r$. 
We could also reach this by quoting the rotation formula. Recall that when we are rotating a point $(u,v)$ about the origin through an angle $\phi$, we multiply the vector by a certain matrix.  You can think of this post as being, in particular, a derivation of the rotation formula.
A: Another way to derive this answer it to use complex coordinates. Lets assume the centre of the circle is in $c=a+b i$ and the initial point in $z_0=x_0+y_0$ which can be written as 
$$z_0=c+r e^{i\phi},$$ 
where $r$ is the radius of the circle and $\phi$ is the initial angle. 
The angle over which you have travelled is equal to $\theta=d/r$ where $d$ is the distance travelled along the circle in counter clockwise direction. The new point $z_1=x_1+i y_1$ is then 
$$
  z_1= c+r e^{i(\phi+\theta)} 
     = c + (z_0-c) e^{i\theta}
     = c + (z_0-c) e^{i d/r}
$$
By writing everything back in Cartesian coordinates we find
$$
  x_1 = \Re(z_1)= a + (x_0-a) \cos(d/r) - (y_0-b)\sin(d/r)
$$
and
$$ 
  y_1 = \Im(z_1)= b + (x_0-a) \sin(d/r) + (y_0-b)\cos(d/r)
$$
