circle envelope tangent in another circle As the picture shows, 

One big circle ,$(0,0)$ ,radius=R, there is a small circle in it, $(m,0)$ ,radius=r . 
G is on the big circle. From G ,we can do two tangent lines about the small circle. Get the points of intersection E and F. line EF has a envelope about G , which seem like a circle.
How to prove it? Since calculating it requires much effort.
Some additional infomation:
By picking special points,I get
the radius of envelope circle is
$$\frac{R \left(m^4-2 m^2 \left(r^2+R^2\right)-2 r^2 R^2+R^4\right)}{\left(m^2-R^2\right)^2}$$
and the circle center
$$\left(\frac{1}{2} \left(\frac{R \left(-m^2+2 m R+2 r^2-R^2\right)}{(R-m)^2}-\frac{R \left(-m^2-2 m R+2 r^2-R^2\right)}{(m+R)^2}\right),0\right)$$

The expository paper Poncelet's theorem by András Hraskó very nicely treats this problem and its relation to Poncelet's Closure Theorem.
In the OP diagram call the big and little circles $e$ and $a$ respectively, and the red circle $c$. (this corresponds to the labels in the paper, see Figures 1,4,5).  Poncelet's Theorem is concerned with scenarios such as Figures 1 and 4, where a polygon is inscribed in $e$ such that the edges touch $a$.  But the paper speculates that Poncelet studied that situation of an inscribed triangle where one of the sides does not touch $a$, and found that the non-touching side generates the envelope of a circle $c$ (Figure 5).  As @Blue has noted in the comments, $c$ is coaxal with the pair $a$ and $e$.
I leave the details to the paper, but can't resist quoting this:

Poncelet's General Theorem: Let $e$ be a circle of a non-intersecting pencil and let $a_1,a_2,\ldots,a_n$ be (not necessarily
  different) oriented circles in the interior of $e$ that belong to the
  same pencil. Starting at an arbitrary point $A_0$ of the circle $e$, the
  points $A_1,A_2,\ldots,A_n$ are constructed on the same circle, such that the
  lines $A_0A_1, A_1A_2, \ldots, A_{n-1}A_n$ touch the circles $a_1,a_2,\ldots, a_n$,
  respectively, in the appropriate direction. It may happen that at the
  end of the construction, we get back to the starting point, that is,
  $A_n=A_0$. The theorem states that in that case, we will always get back
  to the starting point in the $n$-th step, whichever point of $e$ we start
  from. We do not even need to take care to draw the tangents to the
  circles in a fixed order.

The case of $a_1=a_2=\cdots=a_n=a$ gives the classic Poncelet's Closure Theorem.
The case $a_1=a_3=a, a_2=c$ relates to the OP.
