Bézier curve approximation of a circular Arc I would like to know how I can get the coordinates of four control points of a Bézier curve that represents the best approximation of a circular arc, knowing the coordinates of three points of the corresponding circle. I would like at least to know the solution to this problem in the case where two of the known circle points are the two ends of a diameter of the circle.
 A: For a unit semi-circle centered at the origin, the points are $(1,0)$, $(1, \tfrac43)$, $(-1, \tfrac43)$, $(-1,0)$. Translate, rotate, and scale as needed.
If the end-points of the diameter are $\mathbf{P}$ and $\mathbf{Q}$, proceed as follows: 
Let $\mathbf{U}$ be a vector obtained by rotating $\vec{\mathbf{P}\mathbf{Q}}$ through 90 degrees. Then the control points are $\mathbf{P}$, $\mathbf{P} + \tfrac23 \mathbf{U}$, $\mathbf{Q} + \tfrac23 \mathbf{U}$, $\mathbf{Q}$.
Pseudocode is as follows
Vector V = Q - P;
Vector U = new Vector(-A.Y, A.X);   // Perpendicular to PQ
double s = 2.0/3.0;                 // Scale factor
Vector[] controlPoints = { P, P + s*U, Q + s*U, Q };

For general circular arcs, complete details are given in "Good approximation of circles by curvature-continuous Bézier curves", by Tor Dokken, Morten Dæhlen
Tom Lyche, Knut Mørken, Computer Aided Geometric Design Volume 7, Issues 1–4, June 1990, Pages 33-41.
A: You can use the following ways to find the control points of a cubic Bezier curve for approximating a circular arc with end points $P_0$, $P_1$, radius R and angular span A:
Denoting the control points as $Q_0$, $Q_1$, $Q_2$ and $Q_3$, then
$Q_0=P_0$, 
$Q_3=P_1$, 
$Q_1=P_0 + LT_0$ 
$Q_2=P_1 - LT_1$ 
where $T_0$ and $T_1$ are the unit tangent vector of the circular arc at $P_0$ and $P_1$ and $L = \frac{4R}{3}tan(\frac{A}{4})$. 
Please note that above formula will give you a pretty good approximation for the circular arc. But it is not "the best" approximation. We can achieve an even better approximation with more complicated formula for the $L$ value. But for practical purpose, above formula is typically good enough.
A: (I needed a method that was robust in the limit $r\to \infty$, here's what I came up with based on fang's answer.)
We want to approximate an arc through points $A$ and $B$ with radius $r$.  If instead we have 3 points, we can use the side lengths $a, b, c$ of the triangle formed by the points to calculate $r$:
$$\begin{aligned}
s &= \frac{1}{2}(a + b + c) \\
r &= \frac{a b c}{4 \sqrt{s (s - a) (s - b) (s - c)}}
\end{aligned}$$
Let $C_0 = \frac{1}{2}\left(A + B\right)$ be the midpoint of $A$ and $B$.  Let $C = C_0 + D$ be the control point of the quadratic Bézier curve through the points $A$ and $B$ with its midpoint at the midpoint of the arc (where $D$ is perpendicular to $AB$); then
$$|D| = 2 r \left(1 - \cos\left(\sin^{-1}\left(\frac{|A - B|}{2r}\right)\right)\right)$$
Naively computed this has catastrophic cancellation, so I used series expansion, with $x = |A - B|, y = \frac{x}{r}$:
$$|D| = x y \left(\frac{1}{4} + \frac{1}{64} y^2 + \frac{1}{512} y^4 + \frac{5}{16384}y^6 + \frac{7}{131072} y^8 + O(y^{10})\right)$$
If $y$ is not small enough for the series to converge quickly enough, divide the original arc into smaller pieces and retry.
This quadratic Bézier is not a good approximation of a circular arc.  But if we split the original arc in half (the midpoint of the arc is $C_0 + \frac{1}{2} D$), and then find the quadratic control points $C_A$ and $C_B$ for each half as above, they can be modified to give the control points $Q$ of a cubic Bézier curve for the original arc:
$$\begin{aligned}
Q_0 &= A \\
Q_1 &= A + \frac{4}{3}\left(C_A - A\right) \\
Q_2 &= B + \frac{4}{3}\left(C_B - B\right) \\
Q_3 &= B \\
\end{aligned}$$
Diagram:

