How to calculate the coordinates of the center (midpoint) of the arc? I have shape for which I know:
Coordinates of three points (A, B, S) and radius.
How do I calculate the coordinates of the center (midpoint) of the arc between points A and B, please?

Thank you
 A: The ‘centre’ of the arc is its midpoint.  It's fairly easy: the line through $S $ and this midpoint is the bissectrix of the angle $\widehat{ASB}$, and as  you have an isosceles triangle, it is also the median through $S$ of the triangle.
Therefore, once you have determined  the midpoint $I$ of the segment $[AB]$, the unit directing vector of the median is
$\vec u=\frac{\overrightarrow{SI}}{\|\overrightarrow{SI}\|}$, and the midpoint of the arc is simply the point
$$S+ \text{radius}\cdot \vec u$$
A: Claim:  If you draw a line from $S$ to the midpoint of the arc (call it $M_1=(x_4, y_4)$).  And draw a line from $A$ to $B$ ($\overline{AB}$).  The two lines ($\overline{SM_1}$ and $\overline{AB}$) will intersect at that midpoint of the line from $\overline{AB}$ (call that point $M_2$).
Pf: Draw triangles $\triangle M_2SA$ and $\triangle M_2SB$.  They are congruent by $SAS$ ($M_2S = M_2S$ and $\angle M_2SA \cong \angle M_2SB$ and $SA = SB$) so $AM_2 = BM_2$.
.....
we know:

*

*$M_2$ is the midpoint of $\overline{AB}$


*$S, M_1$ and $M_2$ are all on the same line


*$SA= SB =SM_2$
So we have three sets of equations.  Us them to solve for $x_4, y_4$.

*

*Equation of midpoint


 so $M_2 = (\frac {x_1+x_2}2, \frac {y_1+y_2}2)$.



*Equation of slope


$\frac {y_4-y_3}{x_4-x_3} = \frac {\frac {y_1+y_2}2- y_3}{\frac {x_1+x_2}2-x_3} = \frac {y_4-\frac {y_1+y_2}2}{x_4-\frac {x_1+x_2}2}$



*distance formula


 $\sqrt{(x_1-x_3)^2 + (y_1-y_3)^2} = \sqrt{(x_4-x_3)^2 + (y_4-y_3)^2} = \sqrt{(x_2-x_3)^2 + (y_2-y_3)^2}$

