In a circle with centre O, ABC is a secant intersecting the circle at A and B and DC is a tangent to the circle at D. If DC = 12 cm and BC = 6 cm, then the length of AB is equal to:

The correct option is 18 cm.

Given:
Length of the tangent line segment, DC = 12 cm
Length of the external secant segment, BC = 6 cm

Formula/Property:
According to the Tangent-Secant Theorem (also known as the Intersecting Secant-Tangent Theorem), if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the length of the tangent segment is equal to the product of the lengths of the external secant segment and the entire secant segment.

Mathematically, for a circle with an exterior point C, secant ABC, and tangent DC:
${\text{DC}}^2=\text{BC}\times \text{AC}$

Step-by-step Derivation:

1. Substitute the given values into the formula:
${12}^2=6\times \text{AC}$

2. Calculate the square of 12:
$144=6\times \text{AC}$

3. Solve for the total secant length AC:
$\text{AC}=\frac{144}{6}=24\text{cm}$

4. Since point B lies on the secant segment AC between A and C, we have:
$\text{AC}=\text{AB}+\text{BC}$

5. Substitute the known values of AC and BC to find AB:
$24=\text{AB}+6$
$\text{AB}=24-6$
$\text{AB}=18\text{cm}$

Thus, the length of segment AB is 18 cm.