Line tangent to circle 
A circle with a radius of $2$ units has its center at $(0,0)$. A circle with a radius of $7$ units has its center at $(15,0)$. A line tangent to both circles intersects the $x$-axis at $(x,0)$. What is the value of $x$? Express your answer as a common fraction.

My problem with this question is that there are $4$ such tangent lines, so how do I know which one to pick?

So there is solution using the $Ax+By+C=0$ notation for lines. Given a circle with center $(x_c,y_c)$ and radius $r$ the all the possible tangents are given by the line:
$$ x (\cos\psi) + y (\sin \psi) -(x_c \cos\psi + y_c \sin \psi + r) = 0 $$
In fact, every scalar multiple of the $(A,B,C)$ coefficients also describe the same line and we are going to use this fact below to match the tangent lines for the two circles.
For the two circles the two general tangent lines are:
$$\begin{align}
  x (\cos \psi_1) + y ( \sin \psi_1) - 2 & =0 \\
  x (\cos \psi_2) + y ( \sin \psi_2) - (7+15 \cos \psi_2) & = 0
\end{align}$$
If we multiply the first equation with $\lambda = \frac{7}{2} + \frac{15}{2} \cos\psi_2$ and then subtract it from the second equation you will get
$$ x \left(\cos\psi_2 - \lambda \cos\psi_1 \right) + y \left( \sin\psi_2 - \lambda \sin\psi_1 \right) = 0 $$
This must be true, regardless which point $(x,y)$ along the tangent line is used. So the above is alike a set of equations for the two angles
$$\begin{align} \cos \psi_2 & = \lambda \cos \psi_1 \\ \sin \psi_2 & = \lambda \sin \psi_1 \end{align}$$
This has two solutions, $\lambda=1$ and $\lambda=-1$. The first one gives the outside tangents, and the second the inside tangents. So the two tangents are found by $$ \cos \psi_2 = -\frac{1}{3} \\ \cos \psi_2 = - \frac{3}{5} $$
Solving the second tangent line equation for $x$ using $y=0$ and the angles above gives us
$$ x= \frac{7}{\cos \psi_2} + 15 = \begin{cases} -6 & \lambda=1 \\ \frac{10}{3} & \lambda=-1  \end{cases} $$
