locus of moving circles with changing radius Suppose I a curve, $\gamma(t) = (x(t),y(t),R(t))$, which describes the centroids, $(x(t),y(t))$ and radii, $R(t)$, of an infinite number of circles parameterised by $t\in(a,b)$. I would like to find an equation for the curve(s) that bound the area swept out by the circles, any ideas on how I would do this? 
E.g. If both $(x(t),y(t))$ and $R(t)$ vary linearly, the answer would be two semicircles (based on the first and last circles) with straight lines connecting them (gradient of the line is dependent on the rate of change of $\gamma(t)$).

I think it may be necessary to impose a condition, something like $||\dot{R}(t)||<||\left(\dot{x}(t),\dot{y}(t)\right)||$, to ensure that each circle has at least one point on the bounding curve
 A: This is called finding the envelope of a family of curves.
Suppose the family of curves, parametrized by $t$, is defined by
$$
F_t(x,y)=0
$$
We find a point on the envelope by solving
$$
F_t(x,y)=0\quad\text{and}\quad\frac{\partial F_t(x,y)}{\partial t}=0
$$
simultaneously for a given $t$.

For example, suppose that the family of curves is
$$
(x-at)^2+(y-bt)^2-(R+ct)^2=0\tag{1}
$$
Taking the partial with respect to $t$ yields
$$
-2a(x-at)-2b(y-bt)-2c(R+ct)=0\tag{2}
$$
To solve $(1)$ and $(2)$ simultaneously, let $u=\frac{x-at}{R+ct}$ and $v=\frac{y-bt}{R+ct}$. Then we just need to solve
$$
u^2+v^2=1\quad\text{and}\quad au+bv=-c\tag{3}
$$
which has solutions
$$
(u,v)=\frac{\left(-c(a,b)\pm(b,-a)\sqrt{a^2+b^2-c^2}\right)}{a^2+b^2}\tag{4}
$$
For each of the solutions $(u,v)$ given in $(4)$, we get $x=at+u(R+ct)$ and $y=bt+v(R+ct)$. Eliminating $t$, we get the line
$$
(b+vc)x-(a+uc)y=buR-avR\tag{5}
$$
Thus, the envelope of the family of circles given in $(1)$ is the pair of lines given in $(4)$ and $(5)$.
