Question

question 19
is ab a tangent line?

support your answer below.

Explanation:

Step1: Recall Tangent-Segment Theorem

A tangent to a circle is perpendicular to the radius at the point of contact. So, if \( AB \) is tangent to the circle at \( A \), then \( \triangle OAB \) (where \( O \) is the center) should be a right triangle with \( \angle OAB = 90^\circ \). We can check using the Pythagorean theorem. The diameter is 16, so the radius is \( \frac{16}{2}=8 \)? Wait, no, the length from the center to \( A \) is radius, but here the segment from the center to the other end (let's say \( C \)) is diameter 16, so \( AC = 16 \), and the other segment from \( C \) to \( B \) is 20, and \( AB = 12 \). Wait, actually, the triangle is \( \triangle ABC \) where \( AC = 16 \), \( AB = 12 \), \( BC = 20 \). Let's check if \( 12^2 + 16^2 = 20^2 \).

Step2: Calculate Squares

Calculate \( 12^2 = 144 \), \( 16^2 = 256 \), sum is \( 144 + 256 = 400 \). Now \( 20^2 = 400 \). So \( 12^2 + 16^2 = 20^2 \), which means \( \triangle ABC \) is a right triangle with \( \angle A = 90^\circ \). Since \( AC \) is a diameter (passes through the center), and \( \angle A = 90^\circ \), by the tangent-radius theorem, \( AB \) is tangent to the circle at \( A \).

Answer:

Yes, \( AB \) is a tangent line. Because \( 12^2 + 16^2 = 20^2 \), so \( \triangle ABC \) is right - angled at \( A \), and since \( AC \) is a diameter, \( AB \) is perpendicular to the radius (or diameter) at the point of contact \( A \), so \( AB \) is tangent.