Calculate points on an arc of a circle using center, radius and 3 points on the circle

Question

Given the center, radius, and 3 points on a circle, I want to draw an arc that starts at the first point, passes through the second, and ends at the third by specifying the angle to start drawing and the amount of angle to rotate. To do this, I need to calculate the points on the arc. I want the number of points calculated to be variable so I can adjust the accuracy of the calculated arc. This means I probably need a loop that calculates each point by rotating a little after it has calculated a point. I've read the answer to this question Draw arc with 2 points and center of the circle but it only solves the problem of calculating the angles because I don't know how 'canvas.drawArc' is implemented.

Answer 1

This question has two parts:

1. How to find the arc between two points that passes a third point?
2. How to generate a set of points on the found arc?

Let's start with first part. Given three points A, B and C on the (O, r) circle we want to find the arc between A and C that passes through B. To find the internal angle of the arc we need to calculate the oriented angles of AB and AC arcs. If angle of AB was greater than AC, we are in wrong direction:

Va.x = A.x - O.x;
Va.y = A.y - O.y;
Vb.x = B.x - O.x;
Vb.y = B.y - O.y;
Vc.x = C.x - O.x;
Vc.y = C.y - O.y;

tb = orientedAngle(Va.x, Va.y, Vb.x, Vb.y);
tc = orientedAngle(Va.x, Va.y, Vc.x, Vc.y);

if tc<tb
    tc = tc - 2 * pi;
end


function t = orientedAngle(x1, y1, x2, y2)
    t = atan2(x1*y2 - y1*x2, x1*x2 + y1*y2);
    if t<0
        t = t + 2 * pi;
    end
end

Now the second part. You said:

I probably need a loop that calculates each point by rotating a little after it has calculated a point.

But the question is, how little? Since the perimeter of the circle increases as its radius increase, you cannot reach a fixed accuracy with a fixed angle. In other words, to draw two arcs with the same angle and different radii, we need a different number of points. What we can assume to be [almost] constant is the distance between these points, or the length of the segments we draw to simulate the arc:

segLen = someConstantLength;
arcLen = abs(tc)*r;
segNum = ceil(arcLen/segLen);
segAngle = tc / segNum;
t = atan2(Va.y, Va.x); 

for i from 0 to segNum
    P[i].x = O.x + r * cos(t);
    P[i].y = O.y + r * sin(t);
    t = t + segAngle;
end

Note that although in this method A and C will certainly be created, but point B will not necessarily be one of the points created. However, the distance of this point from the nearest segment will be very small.