Calculate Coordinates on Arc, Based on Time of Day Hopefully someone can help me out with this.
I'm trying to calculate the position of a point on an arc, based on a percentage of distance along the circumference (% time of day). Sidenote - I'm writing a computer program to execute this, it's just the math that's stopping me!
I know the length of the chord, the height of the saggita, and the coordinates of the circle. I've attached an image as a diagram to explain this.
Based on the time of day, from $6$A.M $(0,0)$ to $6$P.M $(1400,0)$, where will the point on the arc be? Given that $12$P.M the arc will be at coordinates $(700,500)$.
Saggita = $500$,
Chord = $1400$

Another image, with the things I know and don't, also better illustrating the problem I'm trying to solve.

 A: Here is a general solution for the situation you describe, pictured below. The question states that the three given points (which I'll call $A,B,C$) are on a circle, and indeed these points determine a unique $\color{blue}{circle}$. I consider the problem of finding where on this circle the sun will appear to be (as a function of time), as seen by someone located at the midpoint $O$ of line segment $AB$. 
NB: $O$ is not the center of the blue circle, but is the center of a $\color{orange}{circle}$ around which the sun will appear (from $O$) to move at (approximately) constant speed. Consequently, the sun will appear to move at varying speeds around the blue circle.


*

*$A=(0,0),\ B=(0,2a),\ C=(a,h)$ (Your question is the case $a=700, h=500$.)

*$AB$ is a line segment of length $2a$ with midpoint $O$.

*$C$ is the point at height $h$ on the perpendicular bisector of $AB$.

*The orange circle has radius $a$ and is centered at $O$.

*$P$ is the point on the orange circle such that $O$P is aligned with the sun.

*The blue circle is the unique circle that passes through points $A,B,C$. Let $R$ denote its radius. (Note $R> a$ if $a>h$.) 

*$P'$ (whose coordinates we want to find) is the point on the blue circle such that $OP'$ is aligned with the sun.



Solution ...


*

*Let $O'$ be the center of the blue circle. By considering the right triangle $AOO'$, $$(R-h)^2 + a^2 = R^2\tag{1}$$ therefore $$R = \frac{a^2+h^2}{2h}\tag{2}$$

*With time $t$ measured in hours since midnight,$$\theta(t) = \frac{\pi}{12}t - \frac{\pi}{2}\quad(6\le t\le 18)\tag{3}$$ (To get this, set $\theta(t) = c_1 t + c_2$, then determine $c_1,c_2$ by requiring $\theta(6) = 0, \theta(18) = \pi$.)

*Let $(x_P,y_P)$ be the coordinates of point $P$. By simple trigonometry, $$\begin{align}x_P =& a(1-\cos\theta)\\ y_P =& a\sin\theta\end{align}\tag{4}$$

*Find the required coordinates $(x,y)$ of point $P'$ by solving two simultaneous equations: The line through $OP$ is described by the equation $$y=\begin{cases}\frac{y_P}{x_P-a}(x-a) && \text{if }x_P\neq a\\ h && \text{if }x_P = a\end{cases}\tag{5}$$ and the blue circle has equation $$(x-a)^2 + ((y+(R-h))^2 = R^2\tag{6}$$  Substituting $y$ from $(5)$ into $(6)$ and solving for $x$, 
$$x=\begin{cases}
\frac{a-ST - \sqrt{(a-ST)^2 - (S^2+1)(a^2+T^2-R^2)}}{S^2+1} &&\text{if } \theta < {\pi\over 2} \\
a         &&\text{if } \theta = {\pi\over 2}\\
\frac{a-ST + \sqrt{(a-ST)^2 - (S^2+1)(a^2+T^2-R^2)}}{S^2+1} &&\text{if } \theta > {\pi\over 2} \\
\end{cases}\tag{7}$$
where
$$\begin{align}
S =& \frac{y_P}{x_P-a}\\
T =& R-h -aS
\end{align}\tag{8}$$

Summarizing: To find the coordinates $(x,y)$ of point $P'$, the steps are
$$t \xrightarrow{Eq(3)}\theta\xrightarrow{Eq(4)}(x_P,y_P)\xrightarrow{Eqs(5,7,8)} (x,y).$$

Example ($a=700, h=500, R=740$, time = 9:37 AM)
$$\begin{align}
t &= 9 + 37/60 \approx 9.61666666666667\\
\theta &\approx 0.946841119206924\\
P(x_P,y_P) &\approx (291.025234053796, 568.101787375508)\\
P'(x,y) &\approx (389.286724718565,431.607966423778) 
\end{align}$$
Here's a picture for $120$ equally-spaced times ($\delta t = 6$ minutes) spanning the $12$-hour period:

NB: As point $P$ moves around the orange circle at constant speed, point $P'(x,y)$ moves around the blue circle at speeds that vary with time, moving faster near noon and slower near 6 AM and 6 PM. This shows up as a change in the spacing of the plotted points. 
