Calculate distance relative to radius with only arc length and arc height provided I am struggling to find help on how I can calculate x given I only have the arc height (h) and arc length L (please see below diagram).
I understand that if I have the arc width/chord length w, then I can work out r by:
r= (h/2) * (w²/8h)
which then may aid my quest to calculate x, but I am also struggling to calculate that. Any help would be most grateful!

 A: With respect to chord AB of the circle O, POQ is a diameter perpendicular to it (cutting it at R). Let $\angle RBQ = \alpha$.

It should be clear that $\angle P = \alpha$ and the red marked angles are all equal to $2 \alpha$.
First, we have to find the relation between arc AB’s length (L) and chord AB’s length (2s) when they both subtend the same central angle $\angle AOB = 4\alpha$:-
If r is the radius of that circle, then $L = 4r \alpha$, and $2s = 2(r \sin (2 \alpha))$
Eliminating r from the above, we have $s = \dfrac {L \sin (2\alpha)}{4 \alpha}$
Since $\tan \alpha = \dfrac {h}{s}$, we have $L \sin (2 \alpha) \tan \alpha – 4 h \alpha = 0$
It can be simplified slightly to $L \sin^2 \alpha – 2 h \alpha = 0$, but is still transcendental. Seek help from WolframAlpha to find $\alpha$. Once it is known, the rest is easy.
A: The relation you gave for $r$  is incorrect. Shall make another sketch to illustrate.
But please note what is given and what is required in this problem
Given 
Arc Length $L$ , width $w$, height $h $
Required
$ x = BC  $ from bottom tangent of your sketch.

$$ h(2r-h) = w^2/4, \quad 2r = h+ \frac{w^2}{4h}$$
that helps finding diameter $2r$
Consider similar triangles $ CBA, CAD $ we have $ CA^2 = CB\cdot CD $
$$ x\, 2r = w^2 $$
Plug in $2r$ to find red line length $x$.
In the above arc length $L$ you indicated by red line was not used. It was not required to solve the geometry, it was extra input parameter given, which can be ignored.
Otherwise keep $L$ and ignore $w$ in an alternate way of working it out.
