Bob,
To figure the diameter of a circle that would fit between these two points I would take the difference between your start and end x coordinate and make that a horizontal leg of a right triangle and take the difference betweeen your y coordinates and make that the vertical leg of that same triangle. If you use those values and the good old calculation from Pathagorus (SP?) you can calculate the length of a hypotenuse for that triangle. That calculated distance is the greatest diameter to fit between your start and end point. In your case the horizontal leg equals 1.1875 and the vertical .3182. If you square both and take the square root of the sum of the two you would have a hypotenuse of 1.2294. In this case the radius of the circle is important and to find the greatest radius that will fit between your two points you will have to divide the hypotenuse by two yielding .6147. Guess what happens to the lengths of the other sides of the triangle. Yep, since trig is about ratio and porportion they halve, too, becomming .1591 and .5938. These will become your I and J values.To go clockwise from start to end using I and J in absolute and the maximum radius of .6147 (1/2 the hypotenuse)the program would go something like this:
G1X0Y2.375
G2X1.1875Y2.0568I.5938J-.1591
If your intent is to go 360 degrees around a circle returning to your start point but passing through your other defined location, as your second post indicates, you will have to make a little code adjustment. Now your program to go from the start point, back to the start point, through the second coordinate, in a clockwise direction will look like this:
G1X0Y2.375
G2I.5938J-.1591
Listing the end point of the circle isn't normally required if it is equal to the start point.Hope that helps.