How calculate a circumscribing circle? A little while ago, I asked this question and received what looks to be a decent answer. Now I'm coming back to this and want to try to implement it. Given the following:

S, V, X and E are points on a 2D plane. I want to calculate the circumcentre and radius of the two circles (so I can calculate their intersections to determine the positions of X). The difficult is that for each triangle SVX and XES I know only one angle and one line - A and B are known and the SV line and SE line are known. Given that information only, I can't figure out how to calculate the circles with my barely remembered high school math.
If someone could point me in the right direction to do this, that would be much appreciated.
 A: Update: I originally misread the question to think that $X$'s position was known. However, there's not enough information to determine the position of $X$ uniquely. Instead, it could be as shown below, or it could instead be this point reflected in the line $SE$. Further details about this issue, as well as using an alternate algebraic method, are explained in this answer. If you can determine which $X$ point you want, then you can use the method shown below to determine the corresponding circle.

You can use that the inscribed angle is half of the central angle that subtends the same arc in the circle, such as shown in the diagram above, and on the right side of the linked Wikipedia article. For example, with $SE$, you have that if $O$ is the center of the circle that both points are on, then $\angle SOE = 2B$. Let $M$ be the midpoint of $SE$. Thus, by drawing the perpendicular bisector of $SE$, it goes through $M$ and $O$, with it splitting the angle at $O$ in half, as shown. Thus, you have $\tan B = \frac{|SM|}{|MO|} \implies |MO| = \frac{|SM|}{\tan B}$. You can thus determine the perpendicular bisector line and use this length to determine where the center of the circle is.
To get the circle radius, call it $r$, note it's given by $r = \frac{|SM|}{\sin B}$. Also, you can use basically the same procedure for the other circle to get its corresponding values.
I'll leave it to you to set up the appropriate equations for the various lines and their lengths to get the coordinates of $O$, as well as those of the center of the other circle.
A: Unfortunately, you don’t have enough information for a unique solution. Knowing only the endpoints of the arc and the inscribed angle, there are two possible arcs. You basically need to know which side of the chord’s extension the unknown point $X$ lies on in order to disambiguate them.  
Algebraically, if you have the endpoiints $\mathbf p_1=(x_1,y_1)$ and $\mathbf p_2=(x_2,y_2)$ and the inscribed angle $\alpha$, then the two arcs are described by the equation $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=\sqrt{(x-x_1)^2+(y-y_1)^2}\sqrt{(x-x_2)^2+(y-y_2)^2}\cos\alpha.\tag{*}$$ This equation is simply the formula for the dot product of two vectors, rearranged slightly. Applied to your example, this produces something like the following illustration:

The point $X$ that you’re trying to find is one of the intersections of the blue and red double arcs, but there is another (besides the point $S$ itself, of course).  
Note, too, that if you square the above equation in order to eliminate the radicals to make solving the system of equations easier, you introduce even more spurious solutions: squaring the equation adds the complementary arcs to the solution set, so the equation is now that of two complete circles. The additional arcs are indicated by dashed lines below: 

In this case, squaring introduces two more intersections.  
As a practical matter, instead of working with equation (*), I think it will be easier to work out the centers and radii of the four circles using the inscribed angle theorem†, as described in another answer, compute pairwise intersections, which reduces to computing some line-circle intersections, and then discard potential solutions based on whatever other criteria you might have.  

† Let $d$ be half the chord length. Then, the two circle centers lie on the chord’s perpendicular bisector at a distance of $d/\tan\alpha$ from its intersection with the chord.
