Circle Enclosed by Three CirclesDate: 08/18/99 at 11:04:15 From: Nathan Sokalski Subject: How to find the enclosed circle I want to know how to find the center and radius of a circle that is enclosed by three other circles that all touch each other at one point. Example: Circle 1 Circle 2 Circle 3 Center X 0 .5 .5 Center Y 0 .5 -.5 Radius (2^2-1)/2 .5 .5 I know how to find a circle using 3 points, but I don't know how to find which one point on each circle the enclosed circle will pass through. Could you please help me? Thank you. Nathan Sokalski Date: 08/18/99 at 13:19:13 From: Doctor Rob Subject: Re: How to find the enclosed circle Thanks for writing to Ask Dr. Math. Let the center of the circle be (h,k). Then you can get equations for h and k from the fact that the distance from (h,k) to the center of each of the three given circles, minus the radius of that circle, is equal to the radius r of the circle sought. This gives you three equations in the three unknowns h, k, and r, of the second degree. Solving these three equations simultaneously will give you the answer you seek. Careful, though, because the process of solving will probably give you extraneous solutions, too, which you have to discard after checking. In the example you give above, the first circle radius you wrote, (2^2-1)/2 = (4-1)/2 = 3/2, gives you a circle that is not tangent to the other two. I assume you meant (sqrt[2]-1)/2 for its radius, which will satisfy the conditions of the problem. Then you get the following equations: sqrt(h^2+k^2) - (sqrt[2]-1)/2 = r sqrt[(h-1/2)^2+(k-1/2)^2] - 1/2 = r sqrt[(h-1/2)^2+(k+1/2)^2] - 1/2 = r This leads to h^2 + k^2 = (r+(sqrt[2]-1)/2)^2 (h-1/2)^2 + (k-1/2)^2 = (r+1/2)^2 (h-1/2)^2 + (k+1/2)^2 = (r+1/2)^2 Subtracting the second from the third, you get k = 0, so h^2 = (r+(sqrt[2]-1)/2)^2 h^2 - h = (r+1/2)^2 - 1/2 Subtracting the second from the first, h = (r+(sqrt[2]-1)/2)^2 - (r+1/2)^2 + 1/2 = (2*r+sqrt[2]/2)*(sqrt[2]-2)/2 + 1/2 = (sqrt[2]-2)*(r-1/2) Substituting this back into either of the last pair of equations gives you a quadratic equation in r: h^2 = (r+(sqrt[2]-1)/2)^2 so (6-4*sqrt[2])*(r-1/2)^2 = (r+(sqrt[2]-1)/2)^2 (6-4*sqrt[2])*r^2 + (-6+4*sqrt[2])*r + (3/2-sqrt[2]) = r^2 + (-1+sqrt[2])*r + (3/4-sqrt[2]/2) (5-4*sqrt[2])*r^2 + (-5+3*sqrt[2])*r + (3/4-sqrt[2]/2) = 0 r^2 + ((1+5*sqrt[2])/7)*r + ((1-2*sqrt[2]/28) = 0 (r-(5-3*sqrt[2])/14)*(r-(-1-sqrt(2))/2) = 0 The second root is negative, and so extraneous. Thus r = (5-3*sqrt[2])/14, h = (2*sqrt[2]-1)/7. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 08/19/99 at 13:15:42 From: Doctor Rob Subject: Re: How to find the enclosed circle As a follow-up to my previous response, a little research reveals that there is a very pretty formula for the radius of the enclosed circle. If the radii of the three mutually tangent circles are r1, r2, and r3, and the radius of the enclosed circle is r, then 1/r = 1/r1 + 1/r2 + 1/r3 + 2*sqrt[1/(r1*r2)+1/(r1*r3)+1/(r2*r3)] In your example, 1/r1 = 2/(sqrt[2]-1) = 2 + 2*sqrt[2] 1/r2 = 2 1/r3 = 2 so 1/r = 2 + 2*sqrt(2) + 2 + 2 + 2*sqrt(4+4*sqrt[2]+4+4*sqrt[2]+4) = 6 + 2*sqrt(2) + 2*sqrt(12+8*sqrt[2]) = 6 + 2*sqrt(2) + 4*sqrt(3+2*sqrt[2]) = 6 + 2*sqrt(2) + 4*(1+sqrt[2]) = 10 + 6*sqrt(2) r = 1/(10+6*sqrt[2]) = (10-6*sqrt[2])/(10^2-2*6^2) = (5-3*sqrt[2])/14 which agrees with the answer I gave earlier. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/