Finding common distance to three points on circles and one input range (cont.) 
Having the said subject above, and the previously asked and solved question. We reached to a quartic equation after solving the 4 given equations.
Then we got an additional condition where point $A$ moves on a known circle [this circle also have points (e,f) and (-e,-f) on it, if this will make any difference], and we couldn't determine the range that point $A$ can move so we can get at least one real value for $N$.
If point $A$ moves to extreme right or left, we don't get any real value for $N$.
What would be the range of A values that will lead to one or more real solutions for the quartic equation of N.
 A: Formulation
I'd use the tangent half-angle formula to express $A$ in terms of a variable $a$ as
$$A=\left(c+L\frac{1-a^2}{1+a^2},d+L\frac{2a}{1+a^2}\right)$$
Do the same for the points on the small circles, $B$ and $-B$ described by parameter $u$:
$$B=\left(e+R\frac{1-u^2}{1+u^2},f+R\frac{2u}{1+u^2}\right)$$
Also you need the point under $A$ where the three segments of length $L$ meet:
$$P=\left(A_x,A_y-R-N\right)=\left(c+L\frac{1-a^2}{1+a^2},d+L\frac{2a}{1+a^2}-R-N\right)$$
Then $N$ is characterized by
$$\lVert P-B\rVert^2=\lVert P+B\rVert^2=N^2$$
These are rational equations, i.e. equations using rational functions due to the division from my tangent half-angle formulation. You could make them polynomial equations by multiplying with the common denominator. In my Sage code below, I express the equation as difference being zero, and do that transformation from rational to polynomial by taking the numerator of the difference.
Using a resultant of those polynomial equations, you can eliminate $u$ from them. The remaining equation will have a solution of multiplicity greater than one for those parameters $a$ where solutions appear or disappear. The discriminant will be zero whenever there exists a multiple solution of $N$. So we can say that whenever the number of solutions changes, the discriminant will be zero. The converse is not necessarily true.
You could find the roots of that discriminant, taken as a polynomial in $a$, to find the potentially relevant positions for $A$. Then you would compute the number of solutions between these positions in order to find the ranges where that count is different from zero.
Computations
It is fairly easy to express the computation in Sage like this:
PR.<a,c,d,e,f,L,R,u,N> = QQ[]
P = vector([c+L*(1-a^2)/(1+a^2), d+L*(2*a)/(1+a^2)-R-N])
B = vector([e+R*(1-u^2)/(1+u^2), f+R*(2*u)/(1+u^2)])
eq1 = ((P-B)*(P-B)-N^2).numerator().resultant(((P+B)*(P+B)-N^2).numerator(), u)
eq1.factor()  # Look at factors to get rid of some noise.
assert eq1.mod(64*R^2*(a^2+1)) == 0
eq2 = eq1 // (64*R^2*(a^2+1))  # Remove spurious factor.
eq3 = eq2.discriminant(N)  # SLOW!!!
eq4 = [i.polynomial(a) for i, _ in eq3.factor()]

However the discriminant computation will take excessive amounts of time and memory. It's taking so long that I decided to send this answer before it had finished for me. Later on I used a workaround to presumably make it a bit faster, or maybe I just waited a bit longer. The resulting polynomial could be split into four distinct factors. Three of them had exponent $2$ and a degree in $a$ of at most $4$, but the most interesting factor had exponent $1$ and degree $20$ in $a$.
Things would be a lot easier if you had concrete numbers, instead of doing all of this symbolically. So my recommendation would be getting a computer algebra system where you can enter all your known values as actual numbers, and then do the computation with only the bare minimum of actually unknown values as variables of the polynomials.
Example
To take a concrete example (based on a construction you shared before):
\begin{align*}
c &= -15 & d &= 14 & L &= 30 \\
e &= 14 & f &= 15 & R &= 12
\end{align*}
For this setup the discriminant had 14 zeros. But only 6 of them actually corresponded to a change in the number of solutions, and those 6 all came from the degree 20 factor of the discriminant:
\begin{align*}
a_1 &= -8.5615180578666 &
a_2 &= -1.1646835723394 \\
a_3 &= -0.8299517239535 &
a_4 &= -0.3442609311901 \\
a_5 &= +0.0167654732809 &
a_6 &= +1.3274725315537
\end{align*}
Each pair denoted a range $[a_{2k-1},a_{2k}]$ where there were real solutions of the original equation.
When I find the time, I'll create some illustrations of these.
Corner case
Note that the tangent half-angle formula is unable to represent a single point on the circle, namely for angle $\pi$. That value would correspond to $\lim_{a\to\infty}$. This would show as the leading coefficient of the discriminant being zero. What I mean is that you can work out what the degree of $a$ in the discriminant should be in a number of ways. The easiest is probably a computations using concrete input numbers in general positions. If for a specific input configuration the degree is less than that, you have a zero leading coefficient and know that $a\to\infty$ i.e. $A=(c-L,d)$ is a solution of the discriminant problem as well, and the number of solutions may change at that point. If you want to avoid all this trouble, just always consider that as a point where the number of solutions might change, since you need to check whether you have zero solutions in a given range anyway, so adding one more range shouldn't be a problem.
