Locus of a midpoint Let $Γ_1$ be a circle of radius $4$, and let $Γ_2$ be a circle of radius $14$. The distance between the centers of $Γ_1$ and $Γ_2$ is $25$. Let $A$ be a variable point on $Γ_1$, let $B$ be a variable point on $Γ_2$, and let $M$ be the midpoint of $AB$. Let $S$ be the set of all possible locations of $M$. Then find the area of $S$. I am getting $81\pi$. See the figure 
With one of my friends, I got this.Let C1 be centered at $(0,0)$ and let C2 be $(25, 0)$. The points that will lie on the boundary of S are
1. Midpoint of (-4, 0) and (11, 0) i.e. (3.5, 0)
2. Midpoint of (4, 0) and (39, 0) i.e. (21.5, 0)
3. Midpoint of the tangents joining C1 and C2
Let the angle that the point of intersection of the tanget to circle C1 and C2 makes with the x axis be $\theta $ then
The point of intersection on C1 and C2 are
  $$(x_1, y_1) = (4\cos{\theta}, 4\sin{\theta})$$
  $$(x_2, y_2) = (25+14\cos{\theta}, 14\sin{\theta})$$
The equation of the tangent is
$$ y = -\frac{x}{\tan{\theta}} + c$$
Putting the above two points in the line equation and eliminating c gives
  $$ 10\sin{\theta} = -\frac{25 + 10\cos{\theta}}{\tan{\theta}}$$
  $$ 10\sin{\theta}\tan{\theta} = -25 - 10\cos{\theta}
   \frac{2}{\cos{\theta}} = -5
   \cos{\theta} = -\frac{2}{5}$$
This gives
  $$ \sin{\theta} = \pm \frac{\sqrt{21}}{5}$$
The two pairs of points of intersection in C1 and C2 are
  $$ (x_1, y_1) = (-\frac{8}{5}, \frac{4\sqrt{21}}{5})$$
  $$ (x_2, y_2) = (\frac{97}{5}, \frac{14\sqrt{21}}{5})$$
and
   $$(x_1, y_1) = (-\frac{8}{5}, -\frac{4\sqrt{21}}{5})$$
  $$ (x_2, y_2) = (\frac{97}{5}, -\frac{14\sqrt{21}}{5})$$
This gives the other two mid points as $(\frac{89}{10}, \frac{9\sqrt{21}}{5})$, $(\frac{89}{10}, -\frac{9\sqrt{21}}{5})$
Using all the mid points obtained and putting them into the ellipse equation
  $$ \frac{(x - x_1)^2}{a^2} + \frac{(y - y_1)^2}{b^2}$$
following are obtained
  $$ x_1 = \frac{25}{2}$$ and $$ y_1 = 0$$
 $ a = 9$ and $b=9$.
 Where is it wrong.($81\pi$ is wrong!) Please help, thanks.

We can write
$$A = A_0 + 2 a (\cos\theta, \sin\theta) \qquad B = B_0 + 2 b (\cos\phi, \sin\phi)$$
where $A_0$ and $B_0$ are the centers of the circles, and $2a$ and $2b$ are the radii. Then
$$M = C_0 + (a\cos\theta + b \cos\phi, a\sin\theta+b\sin\phi)$$
where $C_0 = \frac12(A_0+B_0)$. Note that
$$\begin{align}
|\overline{MC_0}|^2 &= (a\cos\theta+b\cos\phi)^2+(a\sin\theta+b\sin\phi)^2 \\[4pt]
&= a^2 + b^2 + 2 a b \cos(\theta-\phi)
\end{align}$$
so that
$$| a - b | \;\leq\; |\overline{MC_0}| \;\leq\; a+b$$
Therefore, the locus of $M$ is definitely confined to the annulus with outer radius $a+b$ and inner radius $|a-b|$. Once you show that $M$ accounts for all points in that annulus, then the area is
$$\pi (a+b)^2 - \pi(a-b)^2 = 4 a b \pi = \pi\cdot(\text{product of radii})$$
