Circle between parabolas The problem I am working is to find the equation of the circle lying between two parabolas and tangent to both parabolas as shown in the figure.  I have tried to solve this problem using Mathematica by equating the equations of the parabolas with the circle, and the derivative of these equations, but get bogged down in the size of intermediate steps. One help I have found is to equate dx/dy of the equations instead of dy/dz.  This leads to a cubic equation instead of a quartic equation as an intermediate step.

 A: Let the center of the circle be in $(x_0,y_0)$ and its radius $r$. Suppose it intersects the first parabola at $(u,u^2)$ and the second at $(v,v^2+1)$. We should convince ourselves first that $u=v=x_0=0$ is the only solution with $u$ or $v$ equals $0$. 
Then, lets suppose that $x_0 \neq 0$. In this case, the line that pass through $(u,u^2)$ and is normal to the parabola at this point is $y_1 = u^2- \frac{x-u}{2u}$, and for the second parabola is $y_2 = v^2+1 - \frac{x-v}{2v}$. These lines meet in $(x_0,y_0)$, so:
$$\frac{x_0-u}{2u} =  u^2- y_0$$
$$\frac{x_0-v}{2v} = v^2+1 - y_0$$
And also, because the points of contact are also points in the circle, we have
$$(u-x_0)^2 + (u^2-y_0)^2 = r^2$$
$$(v-x_0)^2 + ((v+1)^2-y_0)^2 = r^2$$
So:
$$(u-x_0)^2 \left(1 + \frac{1}{4u^2}\right) = r^2$$
$$(v-x_0)^2\left(1+\frac{1}{4v^2}\right)=r^2$$
We can work with this 4 equations and solve for any of the variables, but we need some parameter. Those are enough for Mathematica or other software to compute the locci of the centers and other interesting features
