Finding the envelope of the family $(x-c)^2+y^2=1+c^2$ I have this family of circles: $(x-c)^2+y^2=1+c^2$. I'm to find the envelope of this family.
Going by what I know, I write $$F(x,y,c)=(x-c)^2+y^2-1-c^2=x^2-2xc+y^2-1=0.$$
Then, $$\frac{\delta F(x,y,c)}{\delta c}=-2x=0.$$
Ideally, I should get a function in $c$ in the second step, and then substitute that value in the original equation to get the envelope. But here, the $c$ cancels. What should I do now?
 A: Idially, you would get a funcion in $c$. But as you can see we're not in an ideal situation. This family of curves has an envelope, but it's a very boring one. 
What you did was correct, so for our envelope we must have $x=0$. We must also have, $(x-c)^2+y^2=1+c^2$, simultaniously. This implies that $c^2+y^2=1+c^2\implies y^2=1$. 
So our envelope is given by $\begin{cases}x=0\\y^2=1\end{cases}$. As I said, this is a very boring envelope. Its only points are $(0,1)$ and $(0,-1)$. You can actually see this happening below. In the picture you see a slider $c$, and the circles $(x-c)^2+y^2=1+c^2$. As you can see clearly, the envelope which belongs to this family is exactly $
\big\{(0,\pm1)\big\}$. 

Extra: The envelope of a family of spheres (or circles) is sometimes called a channel surface or a canal surface. When dealing with a family of spheres $\big(C(t),r(t)\big)$, where both the centres $C(t)$ and the radii $r(t)$ are parameterized by the same parameter $t$, we have that we only get a (nonempty) channel surface if $\Vert C'(t)\Vert^2-\big(r'(t)\big)^2\geq0$ for all $t$ in some interval. If this does not hold, then we immediately know that there will be no channel surface. 
