# Finding the distance between two gears

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.

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Does snugly means the lines through $QP$ and $RS$ are tangent to both circles. And hence $SNP$ and $RMQ$ are not straight lines? Diagram is misleading in this respect. – Eric May 24 '11 at 16:37
@Eric Naslund Sorry, I just updated the details. Yes, thanks for pointing that out. – Xan May 24 '11 at 16:43
Good programs do draw things like this are GeoGebra and CarMetal, where you can define the RELATIONSHIPS between geometrical elements, add labels, etc, and take a screenshot. – heltonbiker Mar 22 '12 at 17:50

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 $\sqrt{l^2 + (r_1-r_2)^2}$.