In the given figure, PAB is a secant and PT is a tangent to the circle from P. If PT = 8 cm, PA = 6 cm and AB = x cm, then the value of x is:

The correct answer is $\frac{14}{3}$ cm.

From the image, we can clearly see a circle with an external point P. From P, two lines are drawn:

1. PT — a tangent touching the circle at point T, with length 8 cm.

2. PAB — a secant that enters the circle at point A and exits at point B, where PA = 6 cm and AB = x cm. The chord AB passes through the interior of the circle.

Step 1: Apply the Secant-Tangent Theorem

When a tangent and a secant are drawn from the same external point, the following relationship holds:

${\text{PT}}^2=\text{PA}\times \text{PB}$

This is because the tangent-secant relationship states: the square of the tangent length equals the product of the two secant segments measured from the external point.

Step 2: Express PB in terms of known quantities

From the figure, point A is where the secant first intersects the circle, and B is where it exits. So:

$\text{PB}=\text{PA}+\text{AB}=6+x$

Step 3: Substitute the known values into the formula

Given: PT = 8 cm, PA = 6 cm, PB = (6 + x) cm

$8^2=6\times (6+x)$

$64=6\times (6+x)$

Step 4: Solve for x

Expand the right-hand side:

$64=36+6x$

Subtract 36 from both sides:

$64-36=6x$

$28=6x$

Divide both sides by 6:

$x=\frac{28}{6}=\frac{14}{3}$

Conclusion: The value of x is $\frac{14}{3}$ cm.