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

Since, AC is a tangent to the circle with center P at point A.

∴ ∠PAB = 90°.

Since, AC is a tangent to the circle with center Q at point C.

∴ ∠QCB = 90°.

In △PAB and △QCB,

⇒ ∠PAB = ∠QCB (Both equal to 90°)

⇒ ∠PBA = ∠QBC (Vertically opposite angles are equal)

⇒ △PAB ~ △QCB.

In right angle △PAB,

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

We know that,

In similar triangles ratio of corresponding sides are equal.

$\Rightarrow \dfrac{PA}{QC} = \dfrac{PB}{QB} \\[1em] \Rightarrow \dfrac{6}{3} = \dfrac{10}{QB} \\[1em] \Rightarrow QB = \dfrac{3 \times 10}{6} \\[1em] \Rightarrow QB = \dfrac{30}{6} \\[1em] \Rightarrow QB = 5 \text{ cm}.$

From figure,

QP = QB + PB = 5 + 10 = 15 cm.

Hence, QP = 15 cm.