Nope. For a given glide speed, it's always a circle, regardless of
how the wind varies with altitude. The way to add non-concentric
circles, is to add the circles as though they were concentric, &
then add the vector offsets, resulting in a displaced circle.
The mathematics is as follows:
Imagine a wind vector function W(h), which returns a vector wind
velocity for altitude "h".
Imagine a gliding aircraft velocity vector function G(a,h) in no
wind, which returns a vector aircraft velocity for heading "a" &
altitude "h". In our discussion so far, we have imagined that this
vector is a constant with respect to altitude, but it turns out it
doesn't matter:
For each tiny increment of time, the distance traveled from the
previous spot, is the vector S = (W(h) + G(a,h))x(the tiny increment
of time). Expressed in calculus terms, it's:
dS = (W(h) + G(a,h)) dt
Integrating, we get the vector S = (integral of W(h) dt) +
(integral of G(a,h) dt)
But (integral of W(h) dt) = where the wind will blow you if you make
no progress in the body of air you are in.
And (integral of G(a,h) dt) = where you will end up if there is no
wind. If "a" is a constant, the integral is a line of constant
length, in the direction of "a"
Note that where you end up is the vector sum of these two integrals:
1. Where the wind blows a similar object that makes no progress in
the air mass; and
2. Where the airplane would glide without any wind, which ends up
being a circle.
As Andrew Sarangan said, a circle displaced by a vector is still a
circle.
OT but important (& I agree with you that the "glide ring is an
interesting toy at best"), and to muddy the water further: The best
glide SPEED changes, depending upon whether or not you are heading
into the wind or away from it. To take your 60kt glide with a 60kt
wind, it's obvious that if you maintain the same glide speed, you
will make zero forward progress. So you must increase the normal
glide speed in order to make the best angle of glide. How much? It
depends upon the particular airplane's aerodynamics (the drag
curve), but a rule of thumb is 1/2 the headwind. That does mean
fly 90kts in your 60kt headwind! Conversely, with a tailwind, to
get the best glide angle, you reduce the glide speed, but NEVER
below the minimum sink speed (typically 5-10 knots above stall in a
light airplane). In fact, if you take the graph of any airplane's
drag curve (expressed as rate of descent as a function of true
airspeed), the minimum sink speed is at the bottom of the drag
curve, & the best glide speed in still air, is found as a line
from the (0,0) point on the graph to the tangent on the drag curve.
The best glide speed in a wind, is found in the same way, but with
the starting point of the tangent line as the headwind or tailwind
component of the wind, on the x-axis of the graph (with that point
being on the negative part of the x-axis for a tailwind).
I remember doing slow-flight in Wisconsin in a Cessna 150 at 3,000
AGL, & going backwards over the ground.
My background in math:
As a junior in high school, I studied calculus on my own. At the
end of my junior year, I asked for permission to take the final exam
of the high-school's advanced placement calculus class for seniors.
As a result, the school had me take 2nd year calculus at the local
junior college, where I was also asked to grade the homework papers
of those taking 1st year calculus (my first real job). When
admitted to Caltech as a freshman, I was given junior status in
math, due to a perfect score on the exam for those transferring in
at the junior level. I graduated with a BS in math (I went on to
major in computer science in grad school).
I've been a CFI for over 50 years, with over 2,000 hours of flight
instruction given. I've been the Chief Flight Instructor at two
Part 141 flight schools. I'm not always right, but I do have a
pretty good understanding of the math & physics involved (that's
another story at Caltech).