Two circles with centres A, B are of radii 6 cm and 3 cm respectively. If AB = 15 cm, find the length of a transverse common tangent to these circles.

The two circles with centres A, B are of radii 6 cm and 3 cm and AB = 15 cm are shown in the figure below:

Given, AB = 15 cm.

Let AP = x, then PB = 15 - x

Considering △ATP and △SBP,

∠T = ∠S (Each are equal to 90°)

∠APT = ∠BPS (Vertically opposite angles are equal)

△ATP ~ △SBP by AA axiom.

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

$\therefore \dfrac{AT}{BS} = \dfrac{AP}{PB} \\[1em] \Rightarrow \dfrac{6}{3} = \dfrac{x}{15 - x} \\[1em] \Rightarrow 3x = 6(15 - x) \\[1em] \Rightarrow 3x = 90 - 6x \\[1em] \Rightarrow 3x + 6x = 90 \\[1em] \Rightarrow 9x = 90 \\[1em] \Rightarrow x = 10.$

∴ AP = 10 cm,

From figure,

PB = AB - AP = 15 - 10 = 5 cm.

Now in right-angled triangle ATP,

$AP^2 = AT^2 + TP^2 \\[1em] 10^2 = 6^2 + TP^2 \\[1em] TP^2 = 100 - 36 \\[1em] TP^2 = 64 \\[1em] TP = \sqrt{64} \\[1em] TP = 8 \text{ cm}.$

Similarly in right angled triangle PSB,

$PB^2 = BS^2 + PS^2 \\[1em] 5^2 = 3^2 + PS^2 \\[1em] PS^2 = 25 - 9 \\[1em] PS^2 = 16 \\[1em] PS = \sqrt{16} \\[1em] PS = 4 \text{ cm}.$

Hence, TS = TP + PS = 8 + 4 = 12 cm.

Hence, the length of a transverse common tangent to these circles are 12 cm.