How can I make a spline out of three points that define a circular arc?

Question

I was looking at the descriptions of various curve techniques here and noticed the mention of circular arcs. I want to implement something like this, but I'm not quite sure what the correct formula to use is. I can easily find formulas for Bezier or Hermite curves, but when I search for circular arc splines or circular arc interpolation I get a lot of ultra technical articles that are far beyond what I need. My hunch is that I take the angle formed between the endpoints and the control point and use that radius, but I'm still a little too wobbly conceptually to be able to guess a formula.

For those who cannot follow the link, the circle data is given by a starting point and end point which lie on the circle as well as a control point which is used to control the angle.

What formula can I use to interpolate over and approximate an arc of a circle given these three points?

Answer 1

Let's assume you want a circular arc from point A to point B, guided by control handle H which is the intersection of the tangents at A & B (and thus equidistant from A & B).

Circles being the bread & butter of geometry, there are a hundred ways to skin this cat - but here's one...

If we define:

midpoint = (A + B)/2;
perpendicular = (-(B - midpoint).y, (B - midpoint).x);

then this can equivalently be expressed with two points A & B and a ratio h, such that

H = midpoint + h * perpendicular;

where values of h > 0 correspond to arcs turning to the right/clockwise from A to B, and < 0 are arcs turning left/counterclockwise (assuming x+ points right and y+ points up - sometimes y is inverted). h = 0 is the degenerate case of a straight line - use linear interpolation if h is very close to zero, as the method is numerically unstable when the radius approaches infinity.

We can use this to compute a center point, radius, start angle, and angular delta between the points:

center = midpoint - perpendicular/h;

radius = Length(A - center);

startAngle = atan2(A - center);

endAngle = atan2(B - center);

angleDelta = endAngle - startAngle;
if(angleDelta * h > 0)
{
   angleDelta -= sign(h) * 2 * PI;
}

Then your formula for a point with interpolation weight t in [0, 1] along the arc is:

angle = startAngle + angleDelta * t;
point = center + radius * (cos(angle), sin(angle));

And the length of the arc is simply r * abs(angleDelta)