The phrase you want to research in general is "envelope of a family of curves". An envelope is a curve that's tangent to each member of the family at some point; in this case, it's the "border" of the shape swept out by your circles (though if your family doesn't extend indefinitely in both directions, you'll want to add circular caps at the top and the bottom).
It happens (as you suspect) that the envelope of your family consists of straight lines, and basic geometry should be able to prove this. However, it could easily have been the case that a more elaborate border curve arises, so I'll talk through your envelope as in this Wikipedia article. Note that the process involves taking "partial derivatives", so is technically an aspect of multi-variate calculus, although knowledge of basic single-variate calculus will do.
First, to reduce the number of parameters involved, let's say that the center moves as speed exactly $1$, and that radius increases at rate $k$. (Effectively, I'm just re-scaling your time parameter and setting $k := b/s$.) Note that, because you have discussed the case of the more-rapidly-increasing radius, we may assume $|k| < 1$.)
Now, at time $t$, the center of your circle is at $(0,t)$, and the radius is $r+kt$, so that
$$x^2 + (y-t)^2 = ( r + k t )^2$$
We define $F$ (so that the above is equivalent to $F = 0$) as
$$F := x^2 + (y-t)^2 - ( r + k t )^2$$
The equation of the envelope arises from setting $F$ and $\partial F/\partial t$ (the partial derivative of $F$ with respect to $t$) simultaneously equal to zero. We already know $F$. The partial derivative we seek is simply the derivative where we treat $x$ and $y$ as if they were constants (just as we do with $k$ and $r$).
$$\frac{\partial F}{\partial t}= 2 ( y - t )(-1) - 2 ( r + k t )(k) = 2 \left( t ( 1 - k^2 ) - ( y + k r ) \right)$$
Solving $\partial F/\partial t = 0$ for $t$ (and recalling $|k|<1$) ...
$$t = \frac{y+kr}{1-k^2}$$
... and substituting into $F = 0$ ...
$$x^2 + \left(y-\frac{y+kr}{1-k^2}\right)^2 - \left( r + k \frac{y+kr}{1-k^2} \right)^2 = 0$$
... yields ...
$$\left( 1 - k^2 \right) \left( x^2 \left( 1 - k^2 \right) - \left(yk+r\right)^2 \right) = 0$$
$$\implies x^2 \left( 1 - k^2 \right) = \left(yk+r\right)^2$$
$$\implies x \sqrt{ 1 - k^2 } = \pm \left(yk+r\right)$$
$$\implies x \sqrt{ 1 - k^2 } \mp y k = \pm r$$
... vindicating your suspicion that the border of the swept area consists of lines.
Specifically, when $k = 0$ (the circle's radius doesn't change as the its center climbs higher), we have a pair of vertical lines at distance $r$ from the $x$-axis:
$$x = \pm r$$
When $k \ne 0$, we have the lines with slope-intercept form
$$y = \pm \frac{\sqrt{1-k^2}}{k} x - \frac{r}{k}$$
The lines cross at $(0,-\frac{r}{k})$, where the circle (at time $t=-r/k$) collapses to a point.
All that said, you are now back to your question of finding an equation for the convex hull of your circles. I'll need to come back to this.
After the comment below, I interprete the question differently, again.
The equstion then seems to be whether there are points $A,B,C$ such that the map
$$P\mapsto |P-A|:|P-B|:|P-C|$$
is injective.
The answer is: No.
If $A,B,C$ are collinear, we have reflection symmetry, hence always at least two solutions to a (valid) given proportion $|P-A|:|P-B|:|P-C|$. Therefore assume $A,B,C$ are not collinear. Assume for the moment that $\Delta ABC$ is not isosceles. Relabel the points so that $\angle CBA$ is one of the acute angles of the triangle.
Then $\angle $ the prolonged ray $BC$ intersects the perpendicular bisector $\ell$ of $AB$ (i.e. the locus of points $P$ with $|P-A|:|P-B|=1$) in some point $D$ that is $\ne B$. The locus of points $P$ with $|P-B|:|P-C|=|D-B|:|D-C|$ is a circle $\odot$ with center on the line $BC$, hence intersects the line $\ell$ in a second point $E$ (apart from $D$). We conclude that
$$ |D-A|:|D-B|:|D-C| = |E-A|:|E-B|:|E-C|$$
(There are many more such pairs, this was just especially easy to construct).
The above construction fails only if the circle $\odot$ degenerates to a point, i.e. if $\Delta ABC$ is isosceles. In that case we can make a minor adjustment to move $D$ a little bit away from $B$ by replacing $\ell$ with the locus of points $P$ with $|P-A|:|P-B|=q$ where $q$ is only approximately $1$; this is a (large) circle that intersects $BC$ not orthogonally, hence this leads to a point $D\ne B$ and then $E\ne D$.
Best Answer
The exterior boundary of the pink area could be obtained by unwinding tightly a rope previously laid about one of the circles (in order to cope with the fact that we take for a while the curve and then its tangent), and symmetrically for the other. The curve obtained in this way is called a circle involute with a classical equation:
$$\begin{cases}x&=&a(\cos(t) + t \sin(t))\\y&=&a(\sin(t) - t \cos(t))\end{cases}$$
when we take the origin in the center of one of the circles.
In fact, we have two arcs of circle involutes that are joined smoothly at the top point of the boundary as can be proven by the given equations.
Remark : an important application of involute of circles is the design of gears.
Edit: here is a Matlab implementation giving the graphic below:
Fig. 1: Initial curve in red. The different blue curves that are featured correspond to 12 different shifts from $-\pi/2$ to $\pi/2$ with a step $\pi/12$.
Fig. 2: If the range of $s$ is extended to $[-2\pi,2\pi]$ for example, the blue curves are the same as before. One gets good approximations by considering (red curves) involutes based on modified circles with centers in $(\pm 2,0)$ and radius $2$.