# Analytical Solution for the Problem of Point Location in Arbitrary Planar Domains
## 6 NRA Crossing and Sense for Circular Arcs
### Parametric Expression for Circular Arcs
Having the circular path segments duly defined and parametrized, the calculation of the NRA intersections is as straightforward as for the linear segment. Hence, to adapt for the circular arc, there must exist one (or two) values for \(t\) that generate an NRA crossing, and let us call them \(t_{c}\):  
\[x_{c}+\mathrm{j}y_{c}=(x_{0}+\mathrm{j}y_{0})+r\mathrm{e}^{[\theta_{1}+t_{c}( \theta_{2}-\theta_{1})]},0\leq t_{c}\leq 1 \tag{52}\]  
or in expanded view:  
\[ \begin{cases}x_{c}=x_{0}+r\cos[\theta_{1}+t_{c}(\theta_{2}-\theta_{1})]&0\leq t _{c}\leq 1\\ y_{c}=y_{0}+r\sin[\theta_{1}+t_{c}(\theta_{2}-\theta_{1})]&0\leq t_{c}\leq 1 \end{cases} . \tag{53}\]  
For the same conditions as before, \(x_{c}\leq 0\) and \(y_{c}=0\) as seen in Section 5, we then have:  
\[ \begin{cases}x_{0}\leq-r\cos[\theta_{1}+t_{c}(\theta_{2}-\theta_{1})]&0\leq t _{c}\leq 1\\ y_{0}=-r\sin[\theta_{1}+t_{c}(\theta_{2}-\theta_{1})]&0\leq t_{c}\leq 1\end{cases} . \tag{54}\]  
However, since the circle can cross the NRA twice, two solutions would be expected for \(t_{c}\) in the second equation of (54); indeed, the total number of solutions in \(\theta\) of an equation of the type \(x=\sin(\theta)\) is given by the following:  
\[\theta= \begin{cases}\arcsin x\pm 2k\pi&k=0,1,2,\dots\\ \pi-\arcsin x\pm 2k\pi&k=0,1,2,\dots\end{cases}  \tag{55}\]  
that, when restricted to the \([-2\pi,+2\pi]\) interval, simplifies into, at most, the following 5 solutions:  
\[\theta= \begin{cases}\arcsin x\\ \arcsin x\pm 2\pi\\ -\arcsin x\pm\pi\end{cases} . \tag{56}\]  
Resuming to the variables from (54), we have  
\[ \begin{cases}t_{c_{1}}=\dfrac{\theta_{1}+\arcsin\frac{y_{0}}{\tau}}{\theta_{1 }-\theta_{2}}\\ t_{c_{2,3}}=\dfrac{\theta_{1}+\arcsin\frac{y_{0}}{\tau}\pm 2\pi}{\theta_{1}- \theta_{2}}\\ t_{c_{4,5}}=\dfrac{\theta_{1}-\arcsin\frac{y_{0}}{\tau}\pm\pi}{\theta_{1}- \theta_{2}}\end{cases}  \tag{57}\]  
where all those potential solutions, besides needing to verify the condition \(0\leq t_{c}\leq 1\), must also verify the first condition of (54), that is, \(x_{0}\leq-r\cos[\theta_{1}+t_{c}(\theta_{2}-\theta_{1})]\). Naturally, at most two of the solutions from (57) are valid (two intersections of the \(x\) axis), but there can be only one, or even no, solutions at all. As just mentioned, both potential solutions (let us name them \(t_{c_{A}}\) and \(t_{c_{B}}\)) must be tested whether they verify expression (54). Nonetheless, there are obvious cases that can be tested beforehand to accelerate the numeric algorithms: one is \(|y_{0}|>r\), resulting in a clear impossibility, and the other case is when \(|y_{0}|=r\), implying that the arc is tangent to the \(x\) axis. For the case of circular arcs, the situation of tangency may be interpreted as either a non-intersection or a double touching of the NRA (entering and leaving the NRA at the same point). The net additional contribution for the calculation of the line integral is, therefore, zero. Hence, if \(|y_{0}|\geq r\), then the circular segment does not affect the intersection counter \(I_{C}\). The exception occurs in case the tangency point (\(|y_{0}|=r\)) happens for \(t_{c}\) being 0 or 1, which is a \"half intersection\", as described next.