Distance Between Points of Tangency
Date: 03/20/99 at 17:50:00 From: Julian Wake Subject: Distance between contact points of circles of differing diameters I homeschool my daughter and have come across an interesting problem. Let's say we have two circles that are each 5mm in diameter, and the centers of the circles are exactly 50mm apart. If we place these circles so that their outside edges contact the outside edge of a 2,000mm diameter circle, the points where the smaller circles touch the larger circle would be very close to 50mm apart, because of the size of the larger circle. Now if we take the same smaller circles (with their centers still 50mm apart) and place them on the outer edge of a circle with a diameter of only 100mm, the contact points would be much closer together because the smaller circles would touch the 100mm circle on their "inner" sides. How can we calculate the distance between the contact points of the two smaller circles on the larger circle? We know it has to do with the radii of the three circles and the length of the arc on the larger circle, but haven't been able to come up with a formula to find this distance. Any insight you can offer would be much appreciated! - Julian
Date: 03/23/99 at 12:02:49 From: Doctor Peterson Subject: Re: Distance between contact points of circles of differing diameters Hi, Julian. This is an interesting problem, and actually surprisingly easy to solve! Let's draw the three circles: ***** +++++++++++ ***** * * +++++ d +++++ * * * +---* +++---------------------++++*---+ * * r\*++----------------------------+*/r * ***++ \ x / ++*** + \ / + + \ / + + \ / + + R \ / R + + \ / + + \ / + + + + + + + + + + + + + + + + ++ ++ +++ ++ +++ ++++ +++++ +++++ +++++++++++ I'm calling the radius of the small circles r, the radius of the large circle R, and the distance between the centers of the small circles d. We're looking for x, the distance between the points of tangency. The first thing to notice is that the points of tangency are on the lines connecting the centers of the small circles to the center of the large circle. (If that isn't obvious to you, take some time to think about how to prove it.) We therefore have a pair of similar isosceles triangles: d +-----------------------------------+ r\ /r +---------------------------+ \ x / \ / \ / R \ / R \ / \ / + We can set up a proportion between the two triangles: x d --- = ----- R R + r From this you can solve for x in terms of R, r, and d: R * d x = ----- R + r Let's play with this, as you did with the picture. Suppose R is much larger than r; then R + r is practically the same as R, and x becomes close to d, as you pointed out. Suppose instead that R = d/2 - r, so that the large circle fits exactly between the small circles. Then x = d - 2r (= 2R), as we would expect. Did you notice that this hardly involves the circles at all? We only had to use triangles, and there's no arc length or trigonometry involved in the answer. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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