In the given figure, PQ is a transverse common tangent to two circles with centres A and B and of radii 5 cm and 3 cm respectively. If PQ intersects AB at C such that CP = 12 cm, calculate AB.

From figure,

AP ⊥ PQ (∵ tangent at a point and radius through the point are perpendicular to each other.)

In right angled triangle PAC.

⇒ CA² = PC² + AP²

⇒ CA² = 12² + 5²

⇒ CA² = 144 + 25

⇒ CA² = 169

⇒ CA = $\sqrt{169}$

⇒ CA = 13 cm.

Considering triangles PAC and BCQ,

⇒ ∠APC = ∠BQC = 90°

⇒ ∠PCA = ∠QCB (Vertically opposite angles are equal)

△PAC ~ △QBC by AA axiom.

Since triangles are similar hence, the ratio of their corresponding sides are equal.

$\dfrac{CA}{CB} = \dfrac{PA}{QB} = \dfrac{5}{3}$

$\dfrac{13}{CB} = \dfrac{5}{3}$

⇒ CB = 7.8 cm.

From figure,

AB = AC + CB = 13 + 7.8 = 20.8 cm

Hence, AB = 20.8 cm.