>I am trying to smooth some intersecting lines, and for that I think I need the formula of a circle that is tangent to this two intersecting lines; now, I realize that there will be infinite circles matching this, but I can or would like to <g> specify the tangent points (the points where the circle touches each line), but I am googling for this formula without much success, and all the links that look promising get blocked by the internet filter, so, does anyone know how can I solve this problem?
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>TIA
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>Hugo

That looks quite tricky. Let me see if I can at least get started.

The general equation for a circle, IIRC, is (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center of the circle, and r is the radius. Of course, you don't know any of these three pieces of data for a start.

If you want to specify a circle with points, you normally would need three points, not all on the same line. Pressumably, you would have to solve the simultaneous equations.

With two lines, and the tangency points, I think you have sufficient information to specify the circle. But specifying ONLY the two points would not be enough. Evidently, any point lying on the circle would pass through a tangent line.

If you have two tangency lines, and the points of tangency, you can draw the normal (i.e. perpendicular) line for each, and see where they intersect. That would give you the center of the circle. For the radius, just get the distince of the center thus found, to any of the tangency points.

If you have the equation of a tangency line, to draw a perpendicular line, you should consider the following: for two perpendicular lines, the product of their slopes is -1 (minus 1). Thus, if one line has a slope of 2, the perpendicular thereof has a slope of -1/2.

Well, I guess that "basically" solves the problem; you will have to do some manipulation with equations for lines. As you may remember, a line can be expressed in several ways, one of them is y = ax + b, where a is the slope and b is the y-intersect.