I have the following problem:
In my class, we did a majorly complicated method to figure this out but I think there is a better way to do this... Here is the exact problem:
A belt fits snugly around the two circular pulleys shown.
Find the distance between the centers of the pulleys. Round to the nearest hundredth.
The method we used required that I create this huge triangle across the page. It didn't sound like the best way to do it. Does anybody have a "none-hacked" way to do it (as in, a more straight forward procedure)?
EDIT
I'm sorry about this, yes, the diagram is misleading, I had to recreate the image in an annoying program. Yes, lines RS and line QP are both tangent to both circles. Sorry about that, I never explained the context of the question.
Assuming that "fits snugly" means that PQ is tangent to both circles, at Q and P respectively: Note that the radii MQ and NP are both perpendicular to PQ (as it is tangent). This means that MQ and NP are parallel. Draw a line through N parallel to PQ, meeting MP at Q', say.
So QPNQ' is a rectangle, and NQ'=PQ=14 (in length). The part of MQ "above" Q', namely MQ', has length MQ - Q'Q = MQ - NP = 5 - 4 = 1.
This means that MN, the distance between the centres, is the hypotenuse of a right-angled triangle MQ'N with MQ'=MQ-NP and Q'N=QP, so $$\text{MN} = \sqrt{(\text{MQ}-\text{NP})^2 + \text{QP}^2} = \sqrt{(5-4)^2 + 14^2} = \sqrt{197} \approx 14.04.$$
In general, if the length of the PQ-like part (distance between points of tangency) is $l$, and the circles have radii $r_1$ and $r_2$, then the distance between the centres is $\sqrt{l^2 + (r_1-r_2)^2}$.