Placement of "good" control points in Bezier curves

Question

I've been working on this problem for awhile now, and haven't been able to come up with a good solution thusfar.

The problem: I have an ordered list of three (or more) 2D points, and I want to stroke through these with a cubic Bezier curve, in such a way that it "looks good." The "looks good" part is pretty simple: I just want the wedge at the second point smoothed out (so, for example, the curve doesn't double-back on itself). So given three points, where should one place the two control points that would surround the second point in the triplet when drawing the curve.

My solution so far is as follows, but is incomplete. The idea might also help communicate the look that I'm after.

Given three points, (x1,y1), (x2,y2), (x3,y3). Take the circle inscribed by each triplet of points (if they are collinear, we just draw a straight line between them and move on). Take the line tangent to this circle at point (x2,y2) -- we will place the control points that surround (x2,y2) on this tangent line.

It's the last part that I'm stuck on. The problem I'm having is finding a way to place the two control points on this tangent line -- I have a good enough heuristic on how far from (x2,y2) on this line they should be, but of course, there are two points on this line that are that distance away. If we compute the one in the "wrong" direction, the curve loops around on itself.

To find the center of the circle described by the three points (if any of the points have the same x value, simply reorder the points in the calculation below):

double ma = (point2.y - point1.y) / (point2.x - point1.x);
double mb = (point3.y - point2.y) / (point3.x - point2.x);
CGPoint c; // Center of a circle passing through all three points.
c.x = (((ma * mb * (point1.y - point3.y)) + (mb * (point1.x + point2.x)) - (ma * (point2.x + point3.x))) / (2 * (mb - ma)));
c.y = (((-1 / ma) * (c.x - ((point1.x + point2.x) / 2))) + ((point1.y + point2.y) / 2));

Then, to find the points on the tangent line, in this case, finding the control point for the curve going from point2 to point3:

double d = ...; // distance we want the point. Based on the distance between
                // point2 and point3.
// mc: Slope of the line perpendicular to the line between
// point2 and c.
double mc = - (c.x - point2.x) / (c.y - point2.y);
CGPoint tp; // point on the tangent line
double c = point2.y - mc * point2.x; // c == y intercept
tp.x = ???; // can't figure this out, the question is whether it should be
            // less than point2.x, or greater than?
tp.y = mc * tp.x + c;
// then, compute a point cp that is distance d from point2 going in the direction
// of tp.

Answer 1

It sounds like you might need to figure out the direction the curve is going, in order to set the tangent points so that it won't double back on itself. From what I understand, it would be simply finding out the direction from (x1, y1) to (x2, y2), and then travelling on the tangent line your heuristic distance in the direction closest to the (x1, y1) -> (x2, y2) direction, and plopping the tangent point there.