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Suppose A and B are the centres of the larger and smaller circles respectively. Suppose that the tangent to circle A is at D and the tangent to circle B is at C.

Consider the quadrilateral ABCD.

AD = 6 cm, BC = 4 cm and AB = 8 cm.

Also angles ADC and DCB are right angles so that AD CB.

Thus ABCD is a right angled trapezium.

Draw a perpendicular from B onto AD to meet it at E. Then BCDE is a rectangle, and ABE is a right angled triangle.

EA = DA - DE = DA - CB = 6 - 4 = 2

and

EB = CD (opp sides of a rectangle),

So, in ABE, AB2 = AE2 + EB2

82 = 22 + EB2

64 = 4 + CD2

So CD = sqrt(60) = 2*sqrt(15) cm = 7.746 cm (approx).

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