Let O be the center of first first with radius 2 cm.
Let C be the center of the second circle.
Draw OC, which we are told is 20 cm.
Draw External Tangent at A on circle O and B on circle C.
Draw AC which we know is 16 cm.
Let P be the intesection of OC and AB.
Then triangle OAP is similar to triangle PCB. *
*)Because <OAP = <PCB = 90 because it is tangent and hence perpendicular with radius. And <AP0 = <CPB because they are vertical angles.