Editing Composite Bézier curve (section)

===Using four curves===
Considering only the 90-degree [[unit circle|unit-circular]] arc in the [[Cartesian coordinate system#Quadrants and octants|first quadrant]], we define the endpoints <math>\mathbf{A}</math> and <math>\mathbf{B}</math> with control points <math>\mathbf{A'}</math> and <math>\mathbf{B'}</math>, respectively, as:

:<math>
\begin{align}
\mathbf{A} & = [0, 1] \\
\mathbf{A'} & = [\mathbf{k}, 1] \\
\mathbf{B'} & = [1, \mathbf{k}] \\
\mathbf{B} & = [1, 0] \\
\end{align}
</math>

From the definition of the cubic Bézier curve, we have:

:<math>\mathbf{C}(t)=(1-t)^3\mathbf{A} + 3(1-t)^2t\mathbf{A'}+3(1-t)t^2\mathbf{B'}+t^3\mathbf{B}</math>

With the point <math>\mathbf{C}(t=0.5)</math> as the midpoint of the arc, we may write the following two equations:

:<math>
\begin{align}
\mathbf{C} &= \frac{1}{8}\mathbf{A} + \frac{3}{8}\mathbf{A'}+\frac{3}{8}\mathbf{B'}+\frac{1}{8}\mathbf{B} \\
\mathbf{C} &= \sqrt{1/2} = \sqrt{2}/2
\end{align}
</math>

Solving these equations for the x-coordinate (and identically for the y-coordinate) yields:

:<math>\frac{0}{8}\mathbf + \frac{3}{8}\mathbf{k}+\frac{3}{8} + \frac{1}{8} = \sqrt{2}/2</math>
:<math>\mathbf{k} = \frac{4}{3}(\sqrt{2} - 1) \approx 0.5522847498</math>

Note however that the resulting Bézier curve is entirely outside the circle, with a maximum deviation of the radius of about 0.00027.
By adding a small correction to intermediate points such as 

:<math>
\begin{align}
\mathbf{A'} & = [\mathbf{k}+0.0009, 1-0.00103] \\
\mathbf{B'} & = [1-0.00103, \mathbf{k}+0.0009] ,
\end{align}
</math>

the magnitude of the radius deviation to 1 is reduced by a factor of about 3, to 0.000068 (at the expense of the derivability of the approximated circle curve at endpoints).