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To find the equation of a circle, we use the standard form of a circle's equation:

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

Where:
- \((h, k)\) is the center of the circle.
- \(r\) is the radius of the circle.

### Given Information:
- The center is \((0, -4)\).
- The circle passes through the point \((-4, 2)\).

### Step 1: Substituting the center \((0, -4)\) into the equation.
The equation becomes:
\[
(x - 0)^2 + (y + 4)^2 = r^2
\]
Simplified:
\[
x^2 + (y + 4)^2 = r^2
\]

### Step 2: Finding the radius \(r\).
We use the fact that the point \((-4, 2)\) lies on the circle to find the radius.

Substitute \((-4, 2)\) into the equation:
\[
(-4)^2 + (2 + 4)^2 = r^2
\]
\[
16 + 6^2 = r^2
\]
\[
16 + 36 = r^2
\]
\[
r^2 = 52
\]

### Step 3: Final Equation of the Circle
Now that we have \(r^2 = 52\), the equation of the circle is:
\[
x^2 + (y + 4)^2 = 52
\]

This is the equation of the circle.

Would you like more details or further clarifications? Here are some related questions for further exploration:

1. How do you derive the general equation of a circle from three points?
2. Can a circle have a negative radius? Why or why not?
3. What happens when the center of a circle is shifted along the axes?
4. How would the equation change if the circle was tangent to the x-axis?
5. How do you determine if a point lies inside, on, or outside the circle?

**Tip:** The distance formula is key when solving problems involving radii or distances between points on a plane!

Find an equation of the circle that has center (0, -4) and passes through (-4, 2).

Learn how to find the equation of a circle with its center at (0, -4) and passing through the point (-4, 2). This problem covers key geometric concepts, including the standard form of a circle's equation and radius calculation. Suitable for students in Grades 9-11.

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