Finding End point of an Arc in Cartesian Coordinates while radius, arc length and one end of Arc is given? 
I want to find the position of a robot using single tire model while rotating. I am assuming robot is moving along a circle. I know its radius, length or arc and starting point of arc. This time arc direction is clockwise but it could also be anti-clock wise. Can someone explain how can I find ?
 A: Thanks @Mick A, @Narasimham and other people for helping. 
As it is robotic motion so in most cases it between (0-180 degrees). I have starting point coordinates, the radius of virtual circle along with robot moving and angle. So I am using simple polar coordinates. to find next point.

                       θ  =  S/r
                       x1 =  r*cos(θ) + x
                       y1 =  r*sin(θ) + y


A: Since I wasn't satisfied with the answers I saw here. Here is my approach.
Let's say we have the current point on the circle $(P_x,P_y)$ and some circle centred at $(C_x,C_y)$ (which is unknown for now).
I know from polar coordinates that
$$\cases{P_x = C_x + r\cos(\phi)\\
P_y = C_y + r\sin(\phi)}\tag{1}$$
If I want to find another point on this circle $D_x,D_y$ I can also say:
$$\cases{D_x = C_x + r\cos(\theta+\phi)\\D_y = C_y + r\sin(\theta+\phi)}\tag{2}$$
Here $\phi$ is the angle to my first point and $\theta$ is the angle between my first and second point.
However, since I only care about relative locations I can set $\phi=0$ and now:
From $(1)$,
$$\cases{P_x = C_x + r\\
P_y = C_y}\tag{1*}$$
From $(2)$,
$$\cases{D_x = C_x + r\cos(\theta)\\
D_y = C_y + r\sin(\theta)}\tag{2*}$$
Then substituting $(1*)$ into $(2*)$ I get:
$$\cases{D_x = P_x - r + r\cos(\theta)\\
D_y = P_y + r\sin(\theta)}$$
where again, this theta can be back computed from the arc length $S$ as
$$\theta = \frac Sr$$
Notice that now I don't care about the location of the circle centre (and it doesn't matter).
A: I assume the circle is centred at $(0,0)$. Firstly, the angle $\theta_0$ from the positive $x$-axis to the line segment $(0,0)$ to $(x,y)$. Since $(x,y)$ can be anywhere on the circle, we allow $-\pi\lt\theta_0\leq\pi$. Then we have,
$$\theta_0 = \begin{cases}
\pi/2,  & \text{if $x=0,\; y\gt 0$} \\
-\pi/2,  & \text{if $x=0,\; y\lt 0$} \\
\tan^{-1}(y/x),  & \text{if $x\gt 0$} \\
\tan^{-1}(y/x)+\pi,  & \text{if $x\lt 0,\;y\gt 0$} \\
\tan^{-1}(y/x)-\pi,  & \text{if $x\lt 0,\;y\lt 0$}. \\
\end{cases}$$
The angle of rotation is $\theta=l/r$. Say we have $l\gt 0$ for anti-clockwise and $l\lt 0$ for clockwise. Then the new coordinates $(x_1,y_1)$ are:
\begin{align}
x_1 &= r\cos(\theta_0+\theta) \\
y_1 &= r\sin(\theta_0+\theta).
\end{align}
A: EDIT1:
Given $\varphi= \theta$ used symbol and arclength we can find circle coordinates.
Assume wlog origin as the start point...instead of $(x,y).$  Also you should know the initial direction, else  you cannot take off to the end or future traveled points $(x_1,y_1)$ of the robot, much less  determine it,  because initial slope is essential to solving this problem. $\theta$ is a variable, $\varphi$ is central angle at point $(x_1,y_1).$ Also projected the distance along arbitrary x-axis
$$h = r \sin \alpha = \dfrac{l \sin \alpha  }{\varphi} \tag 1 $$
should be computed to start with.
$$r=\dfrac{l}{\varphi} \tag 2 $$
Updating  answer. Symbol $ \theta $ used here for counter-clockwise rotation of radius vector $\rho$ around the origin and $\varphi$ is rotation around center of circle.
At any point, the segment length $2h$ subtends angle $\alpha$ at the circumference. Using Law of sines
$$ \dfrac{\rho}{\sin (\theta+\alpha)}=\dfrac{2h}{\sin  \alpha} \tag 3$$

we directly compute
$$ (x_1,y_1)=\dfrac{2h \sin (\theta+\alpha)}{\sin  \alpha  }\cdot(\cos \theta,\sin \theta) \tag4$$
Isometric input
The given data $\varphi,l, r $ are isometric invariants, so it is included in the answer for more instruction. Anglw $ \alpha$ as an arbitrary Euclidean motion. It is supplied as an initial condition for integration.
smax = 25.; al = -0.6; r = 5;
circ = {PH'[s] == 1/r, PH[0] == -ArcTan[3/4], Y'[s] == Sin[PH[s]], 
   X'[s] == Cos[PH[s]], X[0] == 0, Y[0] == 0};
NDSolve[circ, {X, Y, PH}, {s, 0, smax}];
{x[t_], y[t_], ph[s_]} = {X[t], Y[t], PH[s]} /. First[%];
ParametricPlot[{x[s], y[s]}, {s, .0, smax}, PlotLabel -> Kreis, 
 GridLines -> Automatic, AspectRatio -> Automatic, 
 PlotStyle -> {Blue, Thick}]
Plot[Tooltip[{ph[s], x[s], y[s]}], {s, .0, smax}, 
 PlotLabel -> Circle_Coords, GridLines -> Automatic, 
 AspectRatio -> Automatic]

In the above given program any  l=smax  can be set to change an Endpoint.

