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Two circles of radius 4 are externally tangent to each other and are internally tangent to a circle of radius 12 at points A and B.  Find the distance AB.

 

 Dec 24, 2020
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Let  the center of the large circle  be O  and let the  centers of the  the smaller circles be M and  N

 

Connect  OM, ON  and MN

 

OM  = ON  =12 - 4   = 8

MN =  $ + 4   = 8

 

So....we  have an equilatera triangle  OMN.....this means that angle MON  = 60°

 

And   OA   =12  = OB

 

So....by the Law of  Cosines we can find AB  thusly

 

AB^2   =  OA^2  + OB^2  - 2 (OA * OB)  cos (60°)

 

AB^2  = 12^2  + 12^2  -  2(12*12) (1/2)

 

AB^2   = 288  - 144

 

AB^2  =  144   take the positive square root

 

AB  =12

 

We might also see  that  since OA and OB  = 12  and angle AOB  also = 60°, then triangles OAB and OMN  are similar...so  triangle OAB is also equilateral....so AB  =   12

 

cool cool cool

 Dec 24, 2020

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