Have you considered finding the intersections using an implicit form for the circles, $$\frac{x^2}{r^2} + \frac{y^2}{r^2} + ax + by + c = 0?$$ This representation doesn't have any coefficients that diverge as the circle approaches a straight line. To find intersections, you'll have to solve a quadratic equation whose leading coefficient could be zero or arbitrarily close to it, but the alternative form of the quadratic formula should be able to deal with that robustly.
You'll then have to do some jiggery-pokery to figure out whether the intersection points lie within the arcs. If the arc's bending angle is smaller than $\pi$, a projection onto the line joining the endpoints will suffice.
(Disclaimer: While all of this feels like it should work, I haven't analyzed it in any detail. Also, there could still be a problem when the circle is close to a line and you want the longer arc. But I can't imagine that's a case that would turn up in any practical application.)
Update: For a concrete example, here is the equation for a circular arc passing through the three points $(0,0)$, $(0.5, h)$, and $(1,0)$: $$\kappa^2 x^2 + \kappa^2 y^2 - \kappa^2 x - 2\eta y = 0,$$ where $$\begin{align}\kappa &= \frac{8h}{4h^2 + 1}, \\ \eta &= \frac{8h(4h^2-1)}{(4h^2+1)^2}.\end{align}$$ As you can see, the coefficients remain bounded as $h \to 0$.
Update 2: Wait, that equation becomes trivial if $h = 0$, which is bad. We really want something like $x^2/r + y^2/r + ax + by + c,$ i.e. multiply the previous expression through by $r$. Then for the same example, our equation becomes $$\kappa x^2 + \kappa y^2 - \kappa x - 2\eta' y = 0,$$ where $\eta' = (4h^2-1)/(4h^2+1)$. Here are some explicit values.
$h = 1/2$: $$2 x^2 + 2 y^2 - 2 x = 0,$$ $h = 0.01$: $$0.07997 x^2 + 0.07997 y^2 - 0.07997 x + 1.998 y = 0,$$ $h = 0$: $$2 y = 0.$$
By the way, in this format, the linear terms will always be simply $-2(x_0/r)x$ and $-2(y_0/r)y$, where the center of the circle is at $(x_0,y_0)$. As the center goes to infinity but the endpoints remain fixed, these coefficients remain bounded and nonzero (i.e. not both zero).
Since you want a point from the given points (based on your comments to Ross' answer) here is an $O(n \log n)$ time algorithm, based on the idea of rotating by $45^{\circ}$ and using the equivalence with Manhattan Distance. This only works for the 2D plane.
Assume the points are $(p_n, q_n)$. You transform them to $(q_n - p_n, q_n + p_n) = (x_n , y_n)$ say.
Now we will see how to compute the $x$ component of the Manhattan distance for each point in $O(n \log n)$ time.
We will assume that $x_1 \lt x_2 \lt ...\lt x_n$. This can be done by sorting them. Also, the algorithm can be modified to deal with the case when they are not distinct. For simplicity, we assume they are unique.
Define $L_m = \sum_{j=1}^{m} x_j$ and $R_m = \sum_{j=m}^{n} x_j$. Note that $R_m = L_n - L_{m-1}$. (Define $L_{0} = 0$)
Note that the $L_i$ and $R_i$ can be computed in $O(n)$ time, by looping through the array and maintaining the cumulative sum.
Now for a given $x_k$ we have that $\sum_{j=1}^{n} |x_k - x_j| = \sum_{j=1}^{k-1} (x_k - x_j) + \sum_{j=k+1}^{n} (x_j - x_k) = R_{k+1} - L_{k-1} - (n-2k+1)x_k$
(This is where we used the distinctness assumption, but can be modified easily to deal with it).
Similarly do with $y_k$.
Thus we can compute the sum of (manhattan) distances for each point to other points in total $O(n)$ time.
Finding the minimum among them is another $O(n)$ (or just keep track while computing the distances).
Best Answer
If the arcs are concentric, I would translate $c$ and $P$ such that $c\to 0$ and $P\to P'$. Then $d=\min(||P'|-(r-R)|,||P'|-(r+R)|).$