To determine the equation of a circle given its center and radius, you can use the standard form of the equation of a circle:

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

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

Given:
- The center \((h, k)\) is \((12, -10)\).
- The radius \(r\) is 3.

Substitute these values into the standard form equation:

\[
(x - 12)^2 + (y + 10)^2 = 3^2
\]

Simplify the equation:

\[
(x - 12)^2 + (y + 10)^2 = 9
\]

So, the equation of the circle is:

\[
(x - 12)^2 + (y + 10)^2 = 9
\]

Would you like further details or have any questions about this problem?

Here are 8 related questions that expand on this topic:

1. How do you convert the standard form of a circle's equation to the general form?
2. What is the general form of the equation of a circle?
3. How do you determine the center and radius from the general form of a circle's equation?
4. What is the equation of a circle with a center at the origin and radius 5?
5. How can you graph the circle given by the equation \((x - 12)^2 + (y + 10)^2 = 9\)?
6. What are the parametric equations for a circle with center \((h, k)\) and radius \(r\)?
7. How do you find the equation of a circle passing through three given points?
8. What is the locus of points equidistant from a given point \((h, k)\)?

**Tip:** When working with the equation of a circle, always double-check the signs of the center coordinates and ensure the radius is correctly squared in the equation.

Learn how to find the equation of a circle with a center at (12, -10) and a radius of 3 using the standard form of the circle equation. This problem covers key concepts in circle geometry and coordinate geometry, suitable for students in Grades 9-12.

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