Product of Radii of Two Circles
Date: 7/22/96 at 5:40:29 From: Matthew Darwent Subject: Product of Radii of Two Circles I have given up on solving this problem: The length of a common internal tangent to two circles is 7, and a common external tangent is 11. Compute the product of the radii of the two circles. Thanks, Matthew
Date: 7/22/96 at 8:12:56 From: Doctor Anthony Subject: Re: Product of Radii of Two Circles Draw the diagram I describe and follow the calculation as set out below. Let O1, O2 be the two centers distance 'd' apart, and let r1, r2 be the radii of the circles with r2>r1. Join O1, O2 and draw the radii from the two centers to the points of contact with the common tangent of length 11 units. From O1 draw a perpendicular to the radius r2 (just drawn) to meet this radius at point S. Now we have a right angled triangle O1 S O2, with the shorter sides of length 11 and (r2-r1) and the hypotenuse equal to 'd'. Then by Pythagoras: d^2 = 11^2 + (r2-r1)^2 ......(1) We now turn to the transverse tangent of length 7 units. Draw the lines from the centers O1, O2 to meet this tangent. Now from the foot of the r2 radius where it meets the tangent draw a line parallel to 'd' (line joining centers) until it meets the line of r1. This will require you to extend r1 backwards to the other side of O1. Let T be the point where the lines meet. We now have another right angled triangle, with vertices at T and the two points of contact of the tangent with the circles. The shorter sides are of length 7 and (r2+r1), and the hypotenuse is 'd'. Again applying Pythagoras we get: d^2 = 7^2 + (r2+r1)^2 .......(2) Now subtract (2) from (1) and we have: 0 = 11^2 - 7^2 + (r2-r1)^2 - (r2+r1)^2 0 = 121 - 49 + r2^2 - 2r2.r1 +r1^2 - r2^2 - 2r2.r1 - r1^2 0 = 72 - 4r2.r1 4r2.r1 = 72 r2.r1 = 18 And so the product of the radii is 18. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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