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The problem states that a diameter of a circle is the segment from \((-5, 6)\) to \((7, -2)\). To solve this problem in the context of precalculus, we can determine the following:

### 1. **Find the center of the circle:**
The center of the circle is the midpoint of the diameter. The midpoint \((x_m, y_m)\) of a segment with endpoints \((x_1, y_1)\) and \((x_2, y_2)\) is calculated using the midpoint formula:

\[
(x_m, y_m) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)
\]

Plugging in the coordinates of the endpoints:

\[
\left(\frac{-5 + 7}{2}, \frac{6 + (-2)}{2}\right) = \left(\frac{2}{2}, \frac{4}{2}\right) = (1, 2)
\]

So, the center of the circle is \((1, 2)\).

### 2. **Find the length of the diameter:**
The length of the diameter can be found using the distance formula between the two points:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Substituting the given points:

\[
d = \sqrt{(7 - (-5))^2 + (-2 - 6)^2} = \sqrt{(7 + 5)^2 + (-8)^2} = \sqrt{12^2 + (-8)^2} = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}
\]

So, the length of the diameter is \(4\sqrt{13}\).

### 3. **Find the radius:**
The radius \(r\) is half the length of the diameter:

\[
r = \frac{d}{2} = \frac{4\sqrt{13}}{2} = 2\sqrt{13}
\]

### 4. **Write the equation of the circle:**
The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is:

\[
(x - h)^2 + (y - k)^2 = r^2
\]

Substituting the center \((1, 2)\) and radius \(2\sqrt{13}\):

\[
(x - 1)^2 + (y - 2)^2 = (2\sqrt{13})^2
\]

Simplifying further:

\[
(x - 1)^2 + (y - 2)^2 = 4 \times 13 = 52
\]

So, the equation of the circle is:

\[
(x - 1)^2 + (y - 2)^2 = 52
\]

### Summary:
- **Center of the circle:** \((1, 2)\)
- **Radius:** \(2\sqrt{13}\)
- **Equation of the circle:** \((x - 1)^2 + (y - 2)^2 = 52\)

Would you like to explore any specific part of this problem further or need any additional details?

### Questions for Further Exploration:
1. How do you calculate the midpoint of a line segment?
2. What is the distance formula and how is it derived?
3. How can you find the equation of a circle given its center and a point on the circle?
4. Can the diameter formula be used to find the radius directly?
5. How would you find the circumference of the circle from this information?
6. What is the significance of the radius in the equation of a circle?
7. How can the equation of a circle be transformed into general form?
8. What does the term \( \sqrt{13} \) represent in the context of the radius?

### Tip:
When solving problems involving circles, always start by identifying the center and radius as these are the most crucial parameters in determining the equation of the circle.

Learn how to find the center, radius, and equation of a circle given the endpoints of its diameter (-5, 6) and (7, -2). This problem covers key geometric concepts including the midpoint formula, distance formula, and the derivation of the circle's equation from its diameter. Suitable for high school students studying precalculus or geometry.

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