I have this problem, when I am given a point A an an XY plane, and I need to find the coordinates of a point B that is of a constant distance of my point A, and my OAB angle is fixed (O being the origin). I need to eventually create a grid in the x-y plane, placing my A in all of the grid points, and need to retireve my B coordinates.
I have made a quick Paint diagram of my problem (sorry for my low artistic skills), where I indicated the position vector of A with a black line - I assume to know this, since I am setting up my grid eventually -, the distance between A and B with a red line - the same length in all cases. I did not draw in the angles, but I tried to make it clear that OAB angle is the same in all cases.
I've so far tried the following approaches:
calculating the slope of the AB red line by knowing the slope of the OA line and the constant angle between them. After this, I assumed that b1 = a1 + d(AB)*cos(slope) and b2 = a2 + d(AB)*sin(slope). However, this method failed me, because the AB red line changes its slope between positive and negative values, and this - for some reason - messes up my above approach.
calculating the slope of AB red line again, but this time I tried to intercept it with a circle of radius of the red line around point A. This approach resulted in a tedious expression, and - obviously - two solutions.
Is there an easy, closed (easy to code) expression of finding b1,b2 given a1,a2, d(AB), and OAB angle?
Thanks in advance!