There are two circles which touch each other externally. The radius of the first circle with centre O is 17 cm and radius of the second circle with centre A is 7 cm. BC is a direct common tangent to these two circles, where B and C are points on the circles with centres O and A, respectively. The length of BC is:

$2 \sqrt{119}$ cm

Join A to O and B. Join O to C. Draw AQ perpendicular to OC.

Now, AQ = BC, as they are opposite sides of rectangle AQCB.

⇒ AO = AP + PQ

⇒ AO = 7 + 17 = 24 cm.

⇒ OQ = CO - CQ

⇒ OQ = 17 - 7 = 10 cm

Now, applying pythagoras theorem in \(\Delta \)AOQ

⇒ \( { AO}^{ 2} \) = \( { AQ}^{ 2} \) + \( { OO}^{ 2} \)

⇒ AQ = √(\( { AQ}^{ 2} \) - \( { OO}^{ 2} \))

⇒ AQ = √(\( { 24}^{ 2} \) - \( { 10}^{ 2} \))

⇒ AQ = 2\(\sqrt {119 }\) cm

Therefore, BC is 2\(\sqrt {119 }\) cm as BC = AQ.