I'm trying to fit integral (non-rational) quadratic Bézier curves to circular arcs.

$$ B(t) = (1 - t)^2 P_0 + 2 t (1 - t) P_1 + t^2 P_2 \tag{1} \label{1} $$

Let the angle of the arc be $2\theta$. Assuming the arc starts at

$$P_0 = (r, 0)$$

and ends at

$$P_2 = (r \cos 2\theta, r \sin 2\theta)$$

we intend to find the control point that lies along the angle bisector and thus would be $$ P_1 = (R \cos \theta, R \sin \theta)$$

for some $R$ which is to be solved for. We know the mid point of the arc

$$ B(\frac 1 2) = (r \cos \theta, r \sin \theta) \tag 2 \label 2 $$

Substituting the values of $P_0, P_1, P_2$ and $t = \frac 1 2$ in $\eqref{1}$ and equating values of the $x$ coordinate in $\eqref{1}$ and $\eqref{2}$, we have

$$r \cos \theta = \frac r 4 + \frac {R \cos \theta} 2 + \frac {r \cos 2\theta} 4 $$

Solving for $R$ and simplifying, I could bring it down to

$$ R = \frac {r (4 \cos \theta - \cos 2\theta - 1)} {2 \cos \theta} \tag 3 \label 3 $$

$$ R = r (2 - \frac {\cos 2\theta - 1} {2 \cos \theta}) \tag 4 \label 4 $$

But upon giving the $\eqref 3$ to Wolfrm Alpha it is able to simplify it to

$$ R = r ( 2 - \cos \theta ) $$

but not when $\eqref 4$ is given. I would like to know the identities that were used to simplify it so well; I am unable to arrive it after trying for quite sometime now.



You've been sloppy about the minus sign on the right-hand side. In fact,


  • $\begingroup$ Silly me! Looking at it for too long, I missed the forest for the trees :) Now the answer is obvious, thanks! $\endgroup$ – legends2k Jan 5 '16 at 8:27
  • $\begingroup$ @legends2k: you're welcome, though I think this was probably a case of missing the trees for the forest. $\endgroup$ – symplectomorphic Jan 5 '16 at 9:32

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