In the figure, PAB is a secant and PT is a tangent to the circle. If PT = 5 cm…

Concept

For a tangent and a secant drawn from the same external point, the square of the tangent length equals the product of the external secant segment and the whole secant length.

Thus, if the tangent is PT and the secant meets the circle first at A and then at B, then PT2 = PA × PB, where PB = PA + AB.

Application

1. Let PB be the whole secant length. Using the tangent-secant theorem, PT2 = PA × PB.

2. Substitute PT = 5 cm and PA = 3 cm: 52 = 3 × PB, so 25 = 3 × PB.

3. Hence PB = 25/3 cm.

4. Since PB = PA + AB, AB = PB − PA.

5. Therefore, AB = 25/3 − 3 = 16/3 cm.

Cross-check

Substituting PA = 3 cm and PB = 25/3 cm gives PA × PB = 25 cm2, the same as PT2. The lengths therefore satisfy the tangent-secant relation, and AB = 16/3 cm.