Two circles with centres P and Q of radii 7 cm and 3 cm respectively, touch each other externally at a point A. BC is a direct common tangent to these two circles where B and C are the points on the circles respectively. The length of BC is :

$2\sqrt{21}$ cm

We have,

The radii of both circles = 3 cm and 7 cm

Join P to Q and B. Join Q to C.

 Draw PM ⊥ CQ.

Now PM = BC, as they are opposite sides of rectangle PMBC.

= PQ = 7 + 3 = 10 cm.

= QM = 7 cm – 3cm = 4 cm

⇒ PM =   \(\sqrt {(PQ)^2 - (QM)^2}\) = \(\sqrt {(10)^2 - (4)^2}\) = 2\(\sqrt {21}\)

⇒ PM = BC = 2\(\sqrt {21}\) cm