I have an equilateral triangle with the knowns: coordinates of all the three corners and the circumcircle radius and center coordinates.
Now I want to change the angle of A ( increase or decrease) while holding the coordinates of A and the triangle always staying in the same circumcircle, and find the new coordinates of B and C which can be B1 and C1 as you can see in the following image : how is it possible to do ?
Let's assume that the circle is centered at the origin, for simplicity (origin is $O$ and radius is $R$). Then, we can find the $x$-coordinate of the point $B$. Notice that the triangle $AOB$ is an isosceles triangle.
Draw a horizontal line through $B$ and let's say that the point where this line crosses $x=0$, let's call that point $D$. Due to the isosceles triangle, the angle $DOB$ is twice that of the angle $OAB$, or $$ DOB =2 \times OAB = 2 \times \frac{\alpha}{2} = \alpha $$ ($\alpha$ is the variable angle "A0" in your picture.) Now the distance from $D$ to $B$ can be solved with trigonometry: $$ \sin{DOB} = \frac{DB}{R} \Rightarrow DB = R\sin{\alpha} $$ Therefore, the $x$-coordinate of the point $B$ is just $$ x_B = R\sin{\alpha} $$ The $y$-coordinate is easy to solve from the definition of a circle: $$ x_B^2 + y_B^2 = R^2 \Rightarrow y_B = \sqrt{R^2 - \left(R\sin{\alpha}\right)^2} = R \sqrt{1- \sin^2{\alpha}} $$
I think we're done here.