A Möbius transformation from exterior of discs to an annulus I've attempted to solve the following textbook problem, but now I'm stuck.

Find a Möbius transformation mapping the part of the extended complex plane lying outside the discs $\{z\in\mathbb{C}:|z-5|\leq4\}$ and $\{z\in\mathbb{C}:|z+5|\leq4\}$ onto a domain $V=\{w\in\mathbb{C}:1<|w|<R\}$. Evalute R!

In my attempt of solving this I want to start by mapping the point -9 to 11 to form an annulus containing the exterior of the aforementioned discs, but I don't know how to do this. After that I want to move this annulus to the desired one, and thus obtain R.
I've considered mapping select points to desired points in the domain of V, but I'm uncertain about which points to map where.
I'd like some advice on how to proceed, and feedback about my train of thought. Are there any useful techniques that I haven't used, and that I should keep in mind for similar problems?
 A: Summary: One possible transformation is $T(z) = 2\frac{z+3}{z-3}$, which maps the region in question onto the region $\{w\in\mathbb{C}:1<|w|<4\}$.

Let $T$ be a desired transformation, and let $a = T^{-1}(0)$ and $b = T^{-1}(\infty)$. Then $T(z) = c\frac{z-a}{z-b}$ for some $c\in\mathbb{C}$. From symmetry considerations, we can take $a$ and $b$ to be real. For $T$ to map the circle $C_1 = \{z\in\mathbb{C}:|z+5|\le 4\}$ to a circle on the origin, we must have that $|T(z)|$ is constant on $C_1$. In particular, since $-1\in C_1$ and $-9\in C_1$, it follows that $|T(-1)| = |T(-9)|$. Thus,
$$\left|c\frac{-1-a}{-1-b}\right| = \left|c\frac{-9-a}{-9-b}\right|\implies \left|\frac{1+a}{1+b}\right| = \left|\frac{9+a}{9+b}\right|.$$
Since $T$ is injective, it follows that
$$ \frac{1+a}{1+b} = \frac{T(-1)}{c}\ne\frac{T(-9)}{c} = \frac{9+a}{9+b}.$$
Hence, $\frac{1+a}{1+b}$ and $\frac{9+a}{9+b}$ are distinct real numbers with the same magnitude, so they must be additive inverses of each other. Hence, $\frac{1+a}{1+b} = -\frac{9+a}{9+b}$, and after a few manipulations we obtain the equation $18 + 10a + 10b + 2ab = 0$.
Applying the same logic to the circle $C_2 = \{z\in\mathbb{C}:|z-5|\le 4\}$ yields the relation
$$ \frac{1-a}{1-b} = -\frac{9-a}{9-b}\implies 18 - 10a - 10 b + 2ab = 0.$$
Subtracting the two equations yields $a+b=0$, i.e. $b=-a$. Substituting this into the first equation yields $18 - 2a^2 = 0\implies a^2 = 9\implies a = \pm 3$. Hence, we can take $a=-3$ and $b=3$, yielding $T(z) = c\frac{z+3}{z-3}$. In this case, $C_1$ should be mapped onto a circle of radius $1$ around the origin, and $C_2$ should be mapped onto a circle of radius $R>1$ around the origin, since $T^{-1}(0)$ is in the interior of $C_1$ and $T^{-1}(\infty)$ is in the interior of $C_2$.
For $T$ to map $C_1$ to a circle of radius $1$ around the origin, we need $|T(z)| = 1$ for all $z\in C_1$. In particular, $|T(-1)| = 1$, i.e.
$$\left|c\frac{-1+3}{-1-3}\right| = 1\implies \left|\frac{c}{2}\right| = 1.$$
Hence, we can take $c = 2$. Now, to show that $T$ does map $C_1$ to a circle of radius $1$ around the origin, it suffices to find three points $z_1,z_2,z_3\in C_1$ such that $|T(z_i)| = 1$, since we know that $T(C_1)$ is a circle, and a circle is uniquely determined by any three of its points. From the above calculations, we already know that $|T(-1)| = |T(-9)| = 1$. We can choose the third point to be $z = -5+4i$, and we can check:
$$|T(-5+4i)| = \left|2\frac{(-5+4i)+3}{(-5+4i)-3}\right| = 2\left|\frac{-2+4i}{-8+4i}\right| = 2\frac{\sqrt{2^2+4^2}}{\sqrt{8^2+4^2}} = 1.$$
Thus $T$ maps $C_1$ to $\{w\in\mathbb{C}:|w| = 1\}$. Similar calculations show that $T$ maps $C_2$ to a circle around the origin. To calculate the radius of that circle, it suffices to evaluate $T$ somewhere on $C_2$, e.g. at $z=1$, yielding
$$ R = |T(1)| = \left|2\frac{1+3}{1-3}\right| = 4. $$
Thus, $T(z) = 2\frac{z+3}{z-3}$ maps the circles $C_1$ and $C_2$ to the circles around the origin of radii $1$ and $4$, so it maps the region between the circles (i.e. the region in question) to the annulus around the origin of radii $1$ and $4$.
