AB is the common tangent to both circles as shown in the given figure…

In the given diagram AB is tangent to both the circles

In circle with center C, \(\angle\)CAE = \({90}^\circ\)

In circle with center D, \(\angle\)DBA = \({90}^\circ\)

= \(\angle\)AEC = \(\angle\)BED   (vertically opposite angles)

= So, \(\Delta \)CAE is similar to \(\Delta \)DBC

= \(\frac{CA}{AE}\) = \(\frac{DB}{BE}\)

= \(\frac{4}{3}\) = \(\frac{DB}{15}\)

= DB = \(\frac{4\;×\;15}{3}\) = 20

In \(\Delta \)CAE

= \( { DE}^{ 2} \) = \( { DB}^{ 2} \) + \( { BE}^{ 2} \)

= \( { DE}^{ 2} \) = \( { 20}^{ 2} \) + \( { 15}^{ 2} \)

= DE = \(\sqrt {625 }\) = 25

= Distance between the centers of the circles

= CE + DE = 5 + 25 = 30 cm

Therefore, the distance between the centers of the circles is 30cm.