The question is: What is arc length anyway?
One possible definition is: For $\epsilon>0$, consider all point sequences $a=x_0,x_1,\ldots, x_n=b$ such that all $x_i$ are on the circle (or in any other set $S$ with respect to which we want to measure an arc length) and the distance $d(x_i,x_{i+1})$ between $x_i$ and $x_{i+1}$ is $<\epsilon$. Let $d_\epsilon(a,b)$ be the infimum of $d(x_0,x_1)+d(x_1,x_2)+\ldots +d(x_{n-1},x_n)$ over all such sequences. Then the arc length from $a$ to $b$ can be defined as $\lim_{\epsilon\to 0} d_\epsilon(a,b)$.
By the triangle inequality, $$d(x_0,x_1)+d(x_1,x_2)+\ldots +d(x_{n-1},x_n)\ge d(x_0,x_n)=d(a,b)$$ for all considered point sequences. Hence $d_\epsilon(a,b)\ge d(a,b)$ for all $\epsilon$, and hence the same inequality holds for the limit.
To show that the arc length is in fact strictly greater, pick any point $c$ on the arc between $a$ and $b$, and note that all point sequences with step width $<\epsilon$ have an intermediate point $x_i$ close to $c$ (that is: with $d(x_i,c)<\epsilon$). Hence in the limit we obtain that the arc length from $a$ to $b$ is $\ge d(a,c)+d(c,b)>d(a,b)$, where the last strict inequality follows from the fact that $c$ is not on the straight line segment $ab$.
You already have one good answer, but here are some pointers to a more pedestrian method
easily accessible to a student of trigonometry.
Given the Cartesian coordinates $(x_0,y_0)$ of the center of a circle in
a Cartesian plane, and the coordinates $(x_1,y_1)$ and
$(x_2,y_2)$ of two points on that circle,
you first want to find the direction from the circle's center to each point.
In many computer programming environments you can do this with an atan2 function;
as already explained in an answer to another question,
the formula for the first of the points is
$$ \theta_1 = \mbox{atan2}(y_1 - y_0, x_1 - x_0).$$
Alternatively, using more traditional trigonometry you can write
$$ \theta_1 = \begin{cases}
\tan^{-1} \left(\frac{y_1 - y_0}{x_1 - x_0}\right) & \mbox{if } x_1 - x_0 > 0, \\
\tan^{-1} \left(\frac{y_1 - y_0}{x_1 - x_0}\right) + \pi & \mbox{if } x_1 - x_0 < 0, \\
\frac\pi2 & \mbox{if } x_1 - x_0 = 0 \mbox{ and } y_1 - y_0 > 0, \\
-\frac\pi2 & \mbox{if } x_1 - x_0 = 0 \mbox{ and } y_1 - y_0 < 0, \\
\end{cases}
$$
The reason this is so complicated is that $\tan^{-1}$ gives answers only
in the range $(-\frac\pi2, \frac\pi2),$ which gives you the directions to points
only on a semicircle, not the whole circle, so we need to do something to get the
points on the other semicircle; and of course we cannot do anything at all with
$\frac{y_1 - y_0}{x_1 - x_0}$
when the denominator of that ratio would be zero.
Either of these methods gives you an angle measured in the anti-clockwise sense
from the direction of the positive $x$-axis to the direction from $(x_0,y_0)$ to $(x_1,y_1)$.
Once you have such angles for both points, $\theta_1$ and $\theta_2$,
subtract the angle of the point you are coming from from the angle of the point
you are going to, and this will give you the angle of the anti-clockwise arc
from one point to the other ...
... with one small hitch, which is that the answer should be in the range
$[0, 2\pi)$ since you don't need to go around the circle more than once to reach the
desired point, but subtracting two angles from either of the methods above
will not always be in that range.
So after computing the difference of the angles, $\theta$,
if you don't have $0 \leq \theta < 2\pi$ you "normalize" the
angle by adding or subtracting a whole multiple of $2\pi$ to get a result
$\theta'$ that is in the range $[0, 2\pi)$.
Once you have a normalized angle $\theta'$
(measured in radians, of course!), the anti-clockwise
arc length is $\theta' r$,
To get the clockwise arc length, subtract the anti-clockwise arc from $2\pi r.$
Best Answer
Knowledge of $H$ and knowledge of the radius $r$ are equivalent, because $(r-H)^2+(\frac12 L)^2 = r^2$. (There is a minor technical issue about this quadratic having two solutions for $H$ once you fix $r$, which I will ignore.)
Knowledge of the radius $r$ and knowledge of the central angle $\theta$ subtended by half the arc are equivalent, because $a=2r\theta$.
Finding out $\theta$ is tricky: from \begin{align*} a &= 2r\theta \\ L &= 2r\sin\theta \end{align*} we get $$ \frac La = \frac{\sin\theta}{\theta} $$ and $\theta\in[0,\pi]$. Now, the function on the RHS (called the sinc function) is strictly decreasing on $[0,\pi]$, so there is in fact just one value of $\theta$ in that interval that satisfies this equation; but I don't know how to find it except numerically.