How to find the angle of an arc to draw graphic
I think you have to determine the center C of your circle first. From your drawing, I suppose you want the last black segment to be tangeantial to the circle. Let say this last segment coordinates are P0 (x0, y0) and P1 (x1, y1).

Then the vector is:

vy = y1-y0
vx = x1-x0
with norm vr = sqrt(vy²+vx²)


And the vector vc from P1 to C is (2 choices depending on orientation):

vcy = +R*vx/vr or -R*vx/vr
vcx = -R*vy/vr or +R*vy/vr


then you obtain C coordinates (cx, cy) applying vc to P1. Now, your rotation angle phi must be expressed in the Cxy frame whatever the curve equation you use. The translation from (X, Y) in Cxy to (x, y) in Oxy is:

y = Y+cy
x = X+cx  


Now about Bezier cubic curves, the equations you give assume you are starting at theta0=0 which is not the case on your figure. You can compute the starting angle with arctan2 (implemented on most of languages):

theta0 = arctan(-vcy, -vcx)


Then to plot a large angle arc with Bezier you actually have to cut it in small angle curves (let's say 30° max). So cut the (theta0, theta0+phi) angle into several (a, b) angles. NB: phi is negative on your figure (clockwise).

For every (a, b) angle, your Bezier cubic curve equation in Oxy frame is then:

K= 4/3 * tan( (b-a)/ 4)

start : y = cy+R*sin(a)
        x = cx+R*cos(a)

cp1 : y = cy+R*(sin(a)+K*cos(a))
      x = cx+R*(cos(a)-K*sin(a))

cp2 : y = cy+R*sin(b)- K*cos(b))
      x = cx+R*cos(b)+ K*sin(b))

end : y = cy+R*sin(b)
      x = cx+R*cos(b) 


I did not not check these equations, I hope there is no mistake in.