Calculus word problem involving two circles I've been having difficulty getting started on this word problem for my calculus class. This is how it goes:
There are two circles, both having a radius of one, that meet tangentially at one point. There are two lines which start at the center point of one circle, and are tangent to the second circle. If P1 is the point where the first line meets the second circle, and P2 is the point where the second line meets the second circle, what is the distance between P1 and P2? So far I've been able to visualize this, any ideas?

 A: Hint: When we have a problem about circles, the centres are almost always useful.
Let $A$ be the centre of the left circle, and $B$ the centre of the right circle. Join $A$ and $B$. Join $P_1$ and $P_2$. Let this line segment meet $AB$ at $Y$.
You can compute $AP_1$ by using the Pythagorean Theorem. Now note that triangles $AP_1B$ and $AYP_1$ are similar. 
A: There are basically two ways you can go here:
Classical geometry. Call the circle centers $C_R$ and $C_L$ and draw the lines $C_L C_R$, $C_RP_1$, $C_RP_2$ and $P_1 P_2$.
Now triangle $C_L C_R P_1$ is right and you know the length of the hypotenuse and one cathetus. Use similar triangles to find the length of $P_1 P_2$.
Analytic geometry. Invent a coordinate system, say, with the origin at $C_R$ and $C_L$ at $(-2,0)$. Then if $(x,y)$ are the coordinates of $P_1$, then $y=\sqrt{1-x^2}$. Write out the equation for when $C_L P_1$ and $C_R P_1$ are perpendicular (i.e., have zero dot product), and solve for $x$. Then your desired distance ifs $2y$.

Immediately, the classical strategy looks like it will be the least painful one.

Edit: A slicker classical approach. Draw the circle of radius 1 centered on the point $T$ where the two original circles meet. Due to Thales's theorem it will contain $P_1$ and $P_2$! Therefore triangle $TC_RP_1$ is equilateral with side length $1$, and the $P_1 P_2$ distance is therefore twice the height of such a triangle, or $\sqrt 3$.
