Arc Midpoint Computation
If an origincentered arc of radius $r$ (see Figure) has the ends at $x = a, x = b$ and midpoint at $x = \mu,$ then
\[2 \mu = \pm \sqrt{(r + a)(r + b)} \pm \sqrt{(r  a)(r  b)}\; , \]
where the first radical gets “−” iff the arc has a negative xintercept, and the second radical gets “+” iff the arc has a positive xintercept.
The same relationship holds for yvalues.
Proof
Denote $p=a/r, q=b/r, s_t=(1+t)/2$. Consider angles, $\alpha=\cos^{1}p, \beta=\cos^{1}q$, and $\gamma=cos^{1}(\mu/r)$. There are four cases.
Case 1. The arc does not have $x$intercepts. Then $\gamma=(\alpha + \beta)/2, \mu=r\cos{1 \over 2}(\cos^{1}p + \cos^{1}q)$. Identity $\cos^{1}p + \cos^{1}q = 2\cos^{1}(\sqrt{{s_p}{s_q}}  \sqrt{{s_{p}}{s_{q}}})$ gives $\mu = r(\sqrt{{s_p}{s_q}}  \sqrt{{s_{p}}{s_{q}}})$. Hence, $2\mu = \sqrt{(r+a)(r+b)}  \sqrt{(ra)(rb)}$; 

Case 2.The arc has positive $x$intercept, but does not have negative one. Then $\gamma=\alpha  \beta/2, \mu=r\cos{1 \over 2}(\cos^{1}p  \cos^{1}q)$. Since $\cos^{1}p  \cos^{1}q = 2\cos^{1}(\sqrt{{s_p}{s_q}} + \sqrt{{s_{p}}{s_{q}}})$, we get $\mu = r(\sqrt{{s_p}{s_q}} + \sqrt{{s_{p}}{s_{q}}})$. Hence, $2\mu = \sqrt{(r+a)(r+b)} + \sqrt{(ra)(rb)}$;
Case 3. The arc has negative $x$intercept, but does not have positive one. This part of proof is similar to case 2 with $\gamma = \pi  {1 \over 2}\alpha  \beta$;
Case 4. The arc has two $x$intersepts. This part of the proof is similar to case 1 with $\gamma = \pi  {1 \over 2}(\alpha + \beta)$
Proof for yvalues can be received using the same approach or otherwise. Q.E.D.
Arc midpoint computation was suggested by first attempt shown in [1]. Details of proof involved identities received from [2].
References
 Oleksandr (Alex) G. Akulov (2009). Dot product finds arc midpoint. Math News, U of W. Volume 111, Issue 6, p. 5.
 Gregory V. Akulov (2010). The slope of the angle bisector relationship in applied and theoretical problems. Vinculum, SMTS. Volume 2, Number 1, p. 49.
Copyright © February, 2011 by Oleksandr (Alex) G. Akulov
