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.
+128475 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
