In the given figure, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively. Given that AB = 8 cm, calculate PQ.

Join AP and CQ.

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

In right angled triangle PAB.

$PB^2 = PA^2 + AB^2 \\[1em] \Rightarrow PB^2 = 6^2 + 8^2 \\[1em] \Rightarrow PB^2 = 36 + 64 \\[1em] \Rightarrow PB^2 = 100 \\[1em] \Rightarrow PB = \sqrt{100} \\[1em] \Rightarrow PB = 10 \text{ cm}.$

Considering △PAB and △BCQ,

∠A = ∠C (Each are equal to 90°)

∠ABP = ∠CBQ (Vertically opposite angles are equal)

△PAB ~ △BCQ by AA axiom.

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

$\therefore \dfrac{AP}{CQ} = \dfrac{PB}{BQ} \\[1em] \Rightarrow \dfrac{6}{3} = \dfrac{10}{BQ} \\[1em] \Rightarrow 2 = \dfrac{10}{BQ} \\[1em] \Rightarrow BQ = \dfrac{10}{2} \\[1em] \Rightarrow BQ = 5 \text{ cm}.$

From figure,

PQ = PB + BQ = 10 + 5 = 15 cm.

Hence, the length of PQ = 15 cm.