How to find mid-point of a slant arc? I have these known variables: width (L), left height (H), right height(H1) and radius(R).
I found the coordinates of the midpoint on the arc when H = H1:
$$x = \frac L2$$
$$y = H + \left(R-\sqrt{R^2-\frac{L^2}{4}}\right)$$
However, I could like to know how to find the coordinates of the midpoint of the arc, when H and H1 are different.
The diagram below is a building draw: starting at the coordinate (0,0), then clockwise the point (0,H1) which is the start of the arc over the top with a radius greater than L/2 and with the endpoint of the arc at (L,H). The heights on the two sides are not the same, with $H>H_1$. The sought-after point at the middle of the arc is marked red.

 A: A circumference has the general equation $(x-a)^2+(y-b)^2 = R^2$. If you already know the radious $R$ then you have two unknowns: $a,b$, the coordinates of the center of the circumference.
From your picture you have two points of the circumference: $[0, H1]$ and $[L, H]$. So you can substitute in the general equation and solve to get $a,b$.

Now let's better put the circumference in parametric form:
$$x= a+ R·cos(t)$$
$$y= b+ R·sin(t)$$
where $t$ is the angle from X-axis counter-clockwise.
You can find the angle $t_d$ from point $[0, H1]$ to point $[L, H]$. Add $\pi/2$ to this angle and you get the angle $t_A$ for point $A$ (perpendicular to chord, goes through the center).
Finally, put this angle in the parametric equations.
A: The center of circle is always on perpendicular bisector of BC. I assume that H1, H, L and MA(you marked as R, but in my figure R=DC is the radius of circle I marked length of MA as MA) are known and radius of circle is marked as R.We have:
$B(0, H_1$ and $C(L, H)$
as shown in figure in triangle DMC we have:
$MC^2+(R-MA)^2=DC^2=R^2$ .   .   .     .   .(1)
Where M is midpoint of BC, it's coordinates are:
$x_M=\frac{L}{2}$ and $y_M=\frac{H+H_1}{2}$
$MC=\frac{\sqrt {(x_c-x_B)^2+(y_c-y_B)^2}}{2}$
Knowing MA and finding MC we can find R.
The equation of BC's perpendicular bisector is:
$y-\frac{H+H_1}{2}=\frac{L}{H_1-H}(x-\frac{L}{2})$ .   .   .   .   (2)
This line crosses the circle centered on D at A, the point we want to find it's coordinates, so we need to find the coordinates of D. D is the intersection of a circle centered on B or C with radius R and perpendicular bisector of BC, so we may write:
$(x-x_B)^2+(y-y_B)^2=R^2$   .    .    .    .    (3)
Now we have a system of equation (2) and (3) that gives $x$ and $y$ for point $D(x_D, y_D)$.The equation of circle centered on D is:
$(x-x_D)^2+(y-y_D)^2=R^2$ . . . . . . (4)
Now A is the intersection of this circle with perpendicular bisector of BC. That is  the solution of system of equations (2) and (4) gives $x_A$ and $y_A$.

A: 
Let the midpoint of the arc $(x_t,y_t)$. From the right triangle ABC in the diagram, we have
$$AB=\sqrt{BC^2+AC^2}=\sqrt{(H-H_1)^2+L^2}\tag{1}$$
$$H-H_1 = AB \sin\theta,\>\>\>\>\> L= AB\cos\theta \tag{2}$$
Since D is the midpoint of AB, its coordinates are,
$$x_d=\frac L2, \>\>\>\>\> y_d=\frac {H_1+H}{2}$$
The $x$- and $y$-coordinates of the midpoint T on the arc can be expressed in terms of $(x_d,y_d)$ as,
$$x_t=x_d -DT\sin\theta,\>\>\>\>\>y_t = y_d +DT\cos\theta\tag{3}$$
where,
$$DT = R-OD = R-\sqrt{R^2-\frac{AB^2}{4}}$$
Substitute above DT and (2) into (3) to obtain the coordinates $(x_t,y_t)$,

$$x_t = \frac{L}{2}-\left(R-\sqrt{ R^2-\frac{AB^2}{4}}\right) \frac{H-H_1}{AB}$$
$$y_t = \frac{H_1+H}{2}+\left(R-\sqrt{ R^2-\frac{AB^2}{4}}\right)\frac{L}{AB}$$
where $AB=\sqrt{(H-H_1)^2+L^2}$.

Note that in the special case where $H_1=H$, we have $AB = L$. As expected, the result simplifies to,
$$x_t=\frac L2,\>\>\>\>\> y_t = H + R-\sqrt{R^2-\frac{L^2}{4}}$$
A: 
\begin{align} 
|CD|&=\sqrt{L^2+(h_2-h_1)^2}
,\\
|OE|&=\sqrt{R^2-\tfrac14\,|CD|^2}
,\\
|EX|&=R-\tfrac12\,\sqrt{4R^2-|CD|^2}
,
\end{align} 
Considering the points as complex numbers 
\begin{align}
A&=0,\quad B=L
,\\
C&=L+i\cdot h_2, \quad D=i\cdot h_1
,\\
E&=\tfrac12\,(C+D)
=\tfrac12\,L+i\cdot\tfrac12\,(h_1+h_2)
,\
\end{align} 
and using the fact that rotation a vector $D-C$ by $90^\circ$
in the complex plane is equivalent to multiplication by $i$,
we can find the center of the circle $O$
as it must be located 
at $|OE|$ units from the point $E$
along the line $OE\perp CD$:
\begin{align}
O&=E+\frac{D-C}{|CD|}\cdot i\cdot |OE|
,\\
O&=
\frac{|OE|}{|CD|}\cdot(h_2-h_1)
+\tfrac12\,L
+i\cdot\left(\tfrac12(h_1+h_2)
-\frac{|OE|\cdot L}{|CD|}\right)
.
\end{align} 
And the sought point $X$ is then found as
\begin{align}
X&=O+\frac{E-O}{|OE|}\cdot R
\\
&=\tfrac12\,L-\frac{(h_2-h_1)|EX|}{|CD|}
+i\cdot\left(\tfrac12\,(h_1+h_2)+\frac{L\,|EX|}{|CD|}\right)
.
\end{align}
