write an equation of the circle described. The circle is tangent to the x-axis at (4,0) and has y-intercepts -2 and -8

To write the equation of the circle, we need to determine a few key pieces of information: the center \((h, k)\) and the radius \(r\).

### Step 1: Analyze the given conditions

- The circle is **tangent to the x-axis at (4, 0)**. 
  - This means the point \((4, 0)\) lies on the circle and the circle touches the x-axis at this point.
  - Since the circle is tangent to the x-axis, the distance from the center of the circle to the x-axis (which is \(k\), the y-coordinate of the center) is equal to the radius \(r\). So, we have \(r = |k|\).

- The circle has **y-intercepts at -2 and -8**. 
  - These are points where the circle intersects the y-axis, which occurs when \(x = 0\).
  - At these intercepts, the coordinates are \((0, -2)\) and \((0, -8)\).

### Step 2: Use the y-intercepts

From the y-intercepts, we know the points \((0, -2)\) and \((0, -8)\) are on the circle. The distance between these points gives us the diameter of the circle. The distance between \(-2\) and \(-8\) on the y-axis is:
\[
\text{Distance} = |(-2) - (-8)| = | -2 + 8 | = 6
\]
Thus, the diameter is 6, which means the radius \(r = 3\).

### Step 3: Find the center

Since the circle is tangent to the x-axis at \( (4, 0) \) and the radius is 3, we know the center must be directly above or below the x-axis. The center is vertically above the point \( (4, 0) \) by a distance of 3. Therefore, the center is at \( (4, 3) \).

### Step 4: Write the equation

The general equation of a circle is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Substituting \(h = 4\), \(k = 3\), and \(r = 3\), we get:
\[
(x - 4)^2 + (y - 3)^2 = 9
\]

### Final Answer:
The equation of the circle is:
\[
(x - 4)^2 + (y - 3)^2 = 9
\]

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Would you like more details or have any questions about the steps? Here are 5 related questions for further exploration:

1. How do you determine the center of a circle given two points on the y-axis?
2. What is the relationship between the radius of a circle and its tangency to an axis?
3. How do you find the equation of a circle given its center and a point on the circle?
4. What does the equation of a circle look like when it is tangent to a line or axis?
5. How can you find the equation of a circle using its diameter?

**Tip:** The center of a circle is equidistant from any point on the circle. The distance from the center to a point on the circle is always the radius.

Learn how to find the equation of a circle that is tangent to the x-axis at (4,0) and has y-intercepts at -2 and -8. This problem involves key concepts in analytic geometry, including the equation of a circle, the distance formula, and properties of tangent lines. Suitable for students in Grades 9-12.

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