Calculate the position of the end of a rope wrapped around a pulley I am struggling with a geometry problem, I hope someone more capable than I may be able to point me in the right direction.
I am considering a rope of a known fixed length, wrapped around a pulley of known diameter, with a variable but known position. One end of the rope is connected to a fixed anchor point, and the other end of the rope is constrained to a track that is horizontal, and its position is determined by the fixed length of the rope (under tension) and the pulley position.
i.e. I am trying to calculate the red dimension in the top image. See the below two images, the first a simple representation of the problem, and the second an annotated diagram to provide some labels for discussion.


I can calculate the position of $(x_5, y_5)$ as follows:
$$b = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} $$
$$c = r, \quad\text{where } 2r \text{ is the diameter of the pulley}$$
$$\hat{C} = \sin^{-1}(\frac{c}{b})$$
$$x_5 = b*\sin\left(\tan^{-1}\left(\frac{y_2-y_1}{x_2-x_1}\right)-\hat{C}\right) $$
$$y_5 = b*\cos\left(\tan^{-1}\left(\frac{y_2-y_1}{x_2-x_1}\right)-\hat{C}\right) $$
However, I am then stuck as to what to do next as both the length of the arc of rope wrapped around the pulley and the remaining straight length of pulley (between $\left(x_3,y_3\right)$ and $\left(x_4,y_4\right)$) are unknown.
Is it possible to set up a series of simultaneous equations that can arrive at an arc length or something similar from which the solution can be derived? (not my forte!). The dimensions in the diagrams are completely arbitrary, but can be used to test any solution - the diameter of the circle is 74.13, and the length of the rope is 821.14.
Edit
So I have attempted to do as suggested by @intelligenti pauca below:
$$ e = sqrt((x_3 - x_2)^2 + (y_3 - y_2)^2) $$
$$ \hat{F} = asin(f/e) $$
$$ x_4 = b*sin((tan^{-1}((y_3 - y_2)/(x_3 - x_2)) - \hat{F})) $$
$$ y_4 = b*cos((tan^{-1}((y_3 - y_2)/(x_3 - x_2)) - \hat{F})) $$
$$ d = sqrt((x_4 - x_3)^2 + (y_4 - y_3)^2) $$
$$ s = \frac{r*\pi}{180}*(tan^{-1}((y_3 - y_4)/(x_3 - x_4))-tan^{-1}((y_3 - y_2)/(x_3 - x_2))) $$
The above can be substituted into each other and the below equation.
$$ L = a + d + s $$
Then is it just a case of trying numerical guesses using something like Newton–Raphson?
 A: 
Since the provided diagram is amply annotated, we decided to describe very concisely the calculation steps necessary to determine the horizontal distance $x_3$, which is the only unknown length. For brevity, let $z = x_2 – x_3$ and $y=y_3 - y_2$. Furthermore, to keep thing short and simple, we also assume $\left(x_1, y_1\right) =  \left(0, 0\right)$.
$$l_\mathrm{r} = AH+HG+GB=a+h+d \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\space\space$$
$$b^2 =x_2^2 + y_2^2 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\space\text{Pythagoras theorem applied to }\triangle OPA \space$$
$$a^2 = b^2 – r^2 = x_2^2 + y_2^2 – r^2 \qquad\qquad\qquad\qquad\quad\text{Pythagoras theorem applied to } \triangle AHO$$
$$c^2 = z^2 + y^2 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\space\space \text{Pythagoras theorem applied to } \triangle BQO  $$
$$d^2 = c^2 – r^2= z^2 + y^2 – r^2 \qquad\qquad\qquad\qquad\quad\text{Pythagoras theorem applied to } \triangle OGB $$
$$\theta = \tan^{-1}\left(\frac{a}{r}\right) = \tan^{-1}\left(\frac{\sqrt{x_2^2 + y_2^2 – r^2}}{r}\right) \qquad\text{right angled } \triangle AHO\qquad\qquad\qquad\qquad $$
$$\omega = \tan^{-1}\left(\frac{y_2}{x_2}\right) \quad\qquad\qquad\qquad\qquad\qquad\qquad\text{right angled } \triangle OPA \qquad\qquad\qquad\qquad\space$$
$$\varphi= \tan^{-1}\left(\frac{y}{z}\right)  \qquad\qquad\qquad\qquad\qquad\qquad\quad\quad\text{right angled } \triangle BQO \qquad\qquad\qquad\qquad\space $$
$$\beta = \tan^{-1}\left(\frac{d}{r}\right) = \tan^{-1}\left(\frac{\sqrt{z^2 + y^2 – r^2 }}{r}\right) \qquad\space\text{right angled }\triangle OCB \qquad\qquad\qquad\qquad\space$$
$\small{\measuredangle GOH =}$ $2\pi - \theta - \omega - \phi - \beta$
$$\qquad\space {\small{=}\space} 2\pi - \tan^{-1}\left(\frac{\sqrt{x_2^2 + y_2^2 – r^2}}{r}\right) -\tan^{-1}\left(\frac{y_2}{x_2}\right) -\tan^{-1}\left(\frac{y}{z}\right) - \tan^{-1}\left(\frac{\sqrt{z^2 + y^2 – r^2 }}{r}\right)$$
$$h = r\left\{ 2\pi - \tan^{-1}\left(\frac{\sqrt{x_2^2 + y_2^2 – r^2}}{r}\right) -\tan^{-1}\left(\frac{y_2}{x_2}\right) ) -\tan^{-1}\left(\frac{y}{z}\right) - \tan^{-1}\left(\frac{\sqrt{z^2 + y^2 – r^2 }}{r}\right)\right\}$$
$\space l_\mathrm{r} = \sqrt{x_2^2 + y_2^2 – r^2} + \sqrt{z^2 + y^2 – r^2 }+$
$$r\left\{2\pi - \tan^{-1}\left(\frac{\sqrt{x_2^2 + y_2^2 – r^2}}{r}\right) -\tan^{-1}\left(\frac{y_2}{x_2}\right) ) -\tan^{-1}\left(\frac{y}{z}\right) - \tan^{-1}\left(\frac{\sqrt{z^2 + y^2 – r^2 }}{r}\right) \right\}\tag{1}$$
You need to solve this behemothian nonlinear equation to determine the value of $x_3$. One needs to resort to numerical methods, such as Newton-Raphson method, to do that. So let’s begin.
Using (1), we define the function $f\left(z\right)$ as shown below.
$f\left(z\right)=\sqrt{x_2^2 + y_2^2 – r^2} + \sqrt{z^2 + y^2 – r^2 }-l_\mathrm{r} +$
$$\quad\quad r\left\{2\pi - \tan^{-1}\left(\frac{\sqrt{x_2^2 + y_2^2 – r^2}}{r}\right) -\tan^{-1}\left(\frac{y_2}{x_2}\right) ) -\tan^{-1}\left(\frac{y}{z}\right) - \tan^{-1}\left(\frac{\sqrt{z^2 + y^2 – r^2 }}{r}\right) \right\}$$
The corresponding Newton-Raphson formula is given by
$$z_{i+1} =z_i - \frac{f\left(z_i\right)}{ f’\left(z_i\right)}\quad\text{with}\quad z_0 = \frac{x_2}{2},$$
$$\text{where}\qquad f’\left(z\right)=\frac{z\sqrt{z^2 + y^2 – r^2 }+ry}{z^2 + y^2}.\qquad\qquad\qquad\qquad$$
For the purpose of testing a case proposed by OP, we implemented an iteration procedure using the aforementioned Newton-Raphson formula in an MS-Excel worksheet. This sheet is available for downloading at DropBox. However, we provide this sheet without implying in any way that data and formulae given in it are free of errors. Therefore, if anybody decides to use the results obtained from those data and formulae, that person should do it at his or her own risk.
As you can see, the procedure converges very quickly (i.e. in four iteration steps) to the answer partly thanks to the educated guess we used to start the process. Interested parties can enter data depicting other cases in respective cells with yellow background to verify themselves whether this set of equations and the method are useful or not.
