I need to find origin of circumference which is defined by two points and vertex angle of isosceles triangle:
I've got radius of triangle by deltaAB /2 /math.sin( math.radians( angle /2 ) ), but what is the right equation to find circle origin?
Please disregard red circle from upper image.
You can find point $T$ as $(\frac 12(x_A+x_B),\frac 12(y_A+y_B))$ Then the slope of $TO$ is the negative reciprocal of the slope of $AB$, so is $-\frac {x_B-x_A}{y_B-y_A}$. The line TO is then $y-y_T=-\frac {x_B-x_A}{y_B-y_A}(x-x_T)$ The circle around $A$ with radius $r$ is $(x-x_A)^2+(y-y_A)^2=r^2$. Solve these two simultaneously and you have your answer. There are two solutions. The other will be the other side of AB (upper left in your diagram)