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.
Best Answer
In addition to the answers given on "circle" being the boundary of a "disk" in a 2-dimensional plane: In arbitrary dimensions one usually calls the set of all $x$ in $\mathbb R^n$ with $\|x\|\le 1$ the (closed) $n$-dimensional unit ball and its boundary, the set of all $x$ with $\|x\|=1$, the $(n-1)$-dimensional unit sphere. Here $\|x\|$ denotes the distance from the origin.
So the first figure would be a $1$-dimensional sphere and the second a $2$-dimensional (closed) ball.
The names have their origin in the case $n=3$ where the $3$-ball actually is a solid ball as you think of it and the $2$-sphere is just the surface of the ball.