<s>[INST] How to find center of an arc given start point, end point, radius, and arc direction? Given an arbitrary arc, where you know the following values: start point $(x_0, y_0)$, end point $(x_1, y_1)$, radius ($r$) and arc direction (e.g. clockwise or counterclockwise from start to end), how can I calculate the arc's center?  I know from this previous post (thanks!) that the center lies on the perpendicular bisector between the two points, but don't know how to calculate it.
thanks!
 [/INST] So I gave up with my pen and paper and just put it into my TI-89
a = pointA x
b = pointA y
c = pointB x
d = pointB y

Start with:
$(x-a)^2+(y-b)^2-(x-c)^2-(y-d)^2=0$
Solve for Y
$y=\frac{-(2(a-c)x-a^s-b^2+c^2+d^2)}{2(b-d)}$
Solve for X in  $(x-a)^2+(y-b)^2=r^2$
$x= 
\frac
{- \left( |b-d|\sqrt{4r^2-a^2+2ac-b^2+2bd-c^2-d^2}-(a+c)\sqrt{a^2-2ac+b^2-2bd+c^2+d^2} \right) }
{ 2 \sqrt{a^2-2ac+b^2-2bd+c^2+d^2}}
$
or 
$x= 
\frac
{ |b-d|\sqrt{4r^2-a^2+2ac-b^2+2bd-c^2-d^2}+(a+c)\sqrt{a^2-2ac+b^2-2bd+c^2+d^2} }
{ 2 \sqrt{a^2-2ac+b^2-2bd+c^2+d^2}}
$
</s>