Please refer to the attached image. Is there a formula to calculate the coordinates of the centre of an arc
Given A,B,C with C on the radius
Graphically:
D mid point A-B.
E=Centre of Arc A-B = Intersection Line through D perpendicular to line AB and Ray through A,C
Thanks
Salvo
The arc doesn't seem to be doing anything for us in this picture. I'm assuming you're drawing a circle going through $A$ and $B$ centered at $E$.
Really, all we need is line $AB$, and its perpendicular bisector. Then, we need the line $AC$, and we proceed to find the intersection of $AC$ and $AB$.
Let $A = (a,a')$, $B = (b,b')$, and $C = (c,c')$. Then the line $AB$ is given by $$f(x)= \frac{a'-b'}{a-b}(x-a)+a'$$ and the line $AC$ is given by $$g(x) = \frac{a'-c'}{a-c}(x-a)+a'.$$ The slopes were found by computing $\frac{y-\text{difference}}{x-\text{difference}}$ for each pair of points, then we found the horizontal and vertical shifts needed.
We want the perpendicular bisector of $AB$, which is given by $$h(x) = -\frac{a-b}{a'-b'}(x-f((a+b)/2))) + a'.$$
The lines intersect when $f(x) = h(x)$, so when $$\frac{a'-c'}{a-c}(x-c)+c' = -\frac{a-b}{a'-b'}(x-f((a+b)/2))) + a'.$$ Now, solve for $x$ and plug it back into either $g$ or $h$ and you have your $y$-coordinate for $E$. It'll be messy, but it shouldn't be too tricky.
