Question

in the figure, line segment AB, is tangent to the circle at point A

Answer 1

Answer:

AB = 12 in

Step-by-step explanation:

given a tangent and a secant from an external point to the circle, then the square of the measure of the tangent is equal to the product of the measures of the secant's external part and the entire secant, that is

AB² = BC × BD = 8 × 18 = 144 ( take square root of both sides )

AB = $\sqrt{144}$ = 12

Answer 2

Answer:

AB = 12 in

Step-by-step explanation:

Theorem

When a secant segment and a tangent segment meet at an exterior point, the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.

Secant: a straight line that intersects a circle at two points.

Tangent: a straight line that touches a circle at only one point.

Given:

• Secant segment = BD
• External secant segment = BC
• Tangent segment = AB

$\sf AB^2=BD \cdot BC$

$\implies \sf AB^2=(8+10) \cdot 8$

$\implies \sf AB^2=18 \cdot 8$

$\implies \sf AB^2=144$

$\implies \sf AB=\pm\sqrt{144}$

$\implies \sf AB=\pm 12$

As distance is positive, AB = 12 in