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Explanation:

Let the radius of the third circle be ‘x’ cm.

AF = DE = 4 cm ⇒ AK = DC = (4 – x) cm
 
In ∆ADC, AC2 = AD2 + DC2.
 
(4 + x)2 = 42 + (4 – x)2
 
∴ x = 1.
 
Hence, option (a).

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