# Finding trig solution to locate the center of an arc that intersects a given arc in the upper right quadrant

Circles with colinear (on X axis) centersWorking entirely in the quadrant where x and y are positive, I'm looking for a trig-based solution for something I can easily construct with a compass but can't (yet) find the elegant solution. Given a circle in normal orientation, then picking a point somewhere on that circle in that upper, right quadrant, I seek a way to locate the center of another circle that intersects that chosen point on the first circle, constrained by that second circle center staying on the x axis.

Noobs can't post pix, but it sure would be easier to describe using the labels in this linked pic: original circle of center C and radius CA; point B is target of intersectio by new circles with centers at F and G whose radii are FE and GD respectively; AD = AE & AD < AH (actually, not interested in AD approaching AH, which would push GD off toward infinity; similarly disinterested in angle ACB approaching extremes of zero or pi/2 rads)

• @José Carlos Santos - thank you for the references -- although I don't readily see how the elaborate notation capabilities of mathjax would be helpfully applied to any of my description. That there exists a better way of describing my problem I have absolutely NO DOUBT - and such might easily have required that fancier notation, but I couldn't see a good way to make myself clear without the drawing, which may yet fall short of making myself understood, and such a drawing did not appear (at first glance) to be in the realm of mathjax. Was I wrong in that? – Pale Writer Apr 9 at 17:16
• I never claimed that the drawing is withen the realm of mathjax. It is just that, for instance, $\frac\pi2$ is easier to read than pi/2. – José Carlos Santos Apr 9 at 17:22