What is the length of

\[\overline{AB}\]?
units
A circle centered around point O. Segment O C is a radius of the circle. Point A lies outside the circle, and line A C is a line that could potentially be tangent to circle O. A line segment connects point A to point O and intersects the circle at point B. Line segment A O, line segment O C, and line A C create the triangle A O C. Side A O is broken into two line segments, A B and B O. Segment A C is twenty-four units. Segment O C is seven units.

To find the length of \(\overline{AB}\), we can use the **Power of a Point Theorem**. According to this theorem, if a point lies outside a circle, and a secant and a tangent are drawn from that point, the square of the length of the tangent segment is equal to the product of the lengths of the entire secant segment and its external part.

In this problem:

- Point \(A\) lies outside the circle.
- \(AC\) is the tangent to the circle, and \(AC = 24\) units.
- \(AB\) and \(BO\) are parts of the secant segment \(AO\), where \(BO\) is the radius of the circle, \(OC = 7\) units.

By the Power of a Point Theorem:

\[
AC^2 = AB \times AO
\]

We know that:

- \(AC = 24\) units
- \(OC = 7\) units
- \(BO = 7\) units (since \(BO\) is also a radius)
- \(AO = AB + BO = AB + 7\) units

Substitute the known values into the equation:

\[
24^2 = AB \times (AB + 7)
\]

Simplify:

\[
576 = AB \times (AB + 7)
\]

This expands to:

\[
576 = AB^2 + 7AB
\]

This is a quadratic equation in the form:

\[
AB^2 + 7AB - 576 = 0
\]

To solve this quadratic equation, we can use the quadratic formula:

\[
AB = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here:

- \(a = 1\)
- \(b = 7\)
- \(c = -576\)

Substitute the values:

\[
AB = \frac{-7 \pm \sqrt{7^2 - 4(1)(-576)}}{2(1)}
\]
\[
AB = \frac{-7 \pm \sqrt{49 + 2304}}{2}
\]
\[
AB = \frac{-7 \pm \sqrt{2353}}{2}
\]

Calculating the square root:

\[
AB = \frac{-7 \pm 48.51}{2}
\]

So, the two possible solutions for \(AB\) are:

\[
AB = \frac{41.51}{2} = 20.755 \quad \text{or} \quad AB = \frac{-55.51}{2} = -27.755
\]

Since \(AB\) must be positive, we have:

\[
AB \approx 20.755 \, \text{units}
\]

So, the length of \(\overline{AB}\) is approximately **20.76 units**.

Would you like any more details or explanations? Here are some related questions you might find interesting:

1. How do you apply the Power of a Point Theorem to different geometric configurations?
2. What is the significance of the quadratic formula in solving geometric problems?
3. Can the Power of a Point Theorem be applied if \(A\) lies inside the circle?
4. How does the radius length affect the calculation in this scenario?
5. What are other methods to find the length of a chord in a circle?

**Tip:** Always double-check your work by substituting your solution back into the original equation to ensure accuracy.

Learn how to calculate the length of AB outside a circle using the Power of a Point Theorem. Step-by-step solution with the application of the quadratic equation formula and theorem. Suitable for high school students.

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