Quote Originally Posted by cropwell View Post
Then I would have expected a square root in the answer. Interesting question, but I would have liked to know how the solution was determined (the method, not the equation derivation).

Cheers,

Rob-T
Well, strictly, I suppose, Pythagoras was a means to an end, and it's one reason why there is a bunch of squared terms in the final answer.

I started by sketching the problem, putting in the dimensions as per my first response (above). You can draw a triangle with vertices at the intersection of horiz and vert lines, centre of small circle, and assumed centre of large circle (radius R). Then drop a perpendicular from centre of small circle to horiz line. You have a right-angled triangle with one side = h, and another = R-x.

Write out the equation for the square on the hypotenuse (length y in my sketch): y^2=h^2+(R-x)^2

But from the sketch you can see that y=R-r, so replace y in the Pythogoras equation with (R-r). Multiply it all out and the R^2 terms magically cancel each other out. Then it's just a matter of rearranging for R, which is where we came in. Technically, we should state some assumptions (like r<=x, r<=h) to make sure that we can actually draw the thing. I am also assuming that the large circle is tangent to the small circles on the side of the circle closer to the vertical line.

I would like to say that it was easy, but the real trick is spotting how to get some kind of relationship between the things we know and the one thing that we don't, and I happened to strike lucky with my "draw a triangle" approach.