Locus-Pocus

Let $\Lambda$ be the circle tangent to both $\Gamma_1$ and $\Gamma_2$ . Also let $r_1$ and $r_2$ be the radii of $\Gamma_1$ and $\Gamma_2$ , respectively, and let $x$ be the radius of $\Lambda$ . Finally, let $G_1$ and $G_2$ be the centers of $\Gamma_1$ and $\Gamma_2$ , respectively, and let $L$ be the center of $\Lambda$ .

There are four cases to consider.

Case 1: $\Lambda$ is inside both $\Gamma_1$ and $\Gamma_2$

In this case, $|G_1L-LG_2| = |(r_1-x)-(r_2-x)| = |r_1-r_2|$ , which is constant.

Case 2: $\Lambda$ is outside both $\Gamma_1$ and $\Gamma_2$

In this case, $|G_1L-LG_2| = |(r_1+x)-(r_2+x)| = |r_1-r_2|$ , the same constant as in Case 1 above. Therefore, in both Case 1 and Case 2 , the center $L$ lies on a hyperbola whose foci are $G_1$ and $G_2$ .

Case 3: $\Lambda$ is inside $\Gamma_1$ but outside $\Gamma_2$

In this case, $G_1L+LG_2 = (r_1-x)+(r_2+x) = r_1+r_2$ , which is constant.

Case 4: $\Lambda$ is outside $\Gamma_1$ but inside $\Gamma_2$

In this case, $G_1L+LG_2 = (r_1+x)+(r_2-x) = r_1+r_2$ , the same constant as in Case 3 above. Therefore, in both Case 1 and Case 2 , the center $L$ lies on an ellipse whose foci are $G_1$ and $G_2$ .

I realize that this is not a complete/rigorous proof, but there's enough there to make "an ellipse and a hyperbola" at least the best answer among the choices given!