Find the standard form of the equation for the circle with the following properties.

Center (3,−6)
and radius 6

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

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

where \((h, k)\) is the center of the circle, and \(r\) is the radius.

Given:
- Center: \((h, k) = (3, -6)\)
- Radius: \(r = 6\)

Substituting the values into the formula:

\[
(x - 3)^2 + (y + 6)^2 = 6^2
\]

Simplifying \(6^2\):

\[
(x - 3)^2 + (y + 6)^2 = 36
\]

So, the standard form of the equation of the circle is:

\[
(x - 3)^2 + (y + 6)^2 = 36
\]

Would you like more details on how this was derived or have any other questions? 

Here are some related questions:

1. How do you find the center and radius from the standard form of a circle's equation?
2. What is the general form of a circle's equation?
3. How does the equation of a circle change if the center is at the origin?
4. What are the steps to convert a circle's equation from general to standard form?
5. How do you graph a circle given its equation in standard form?

**Tip:** When the equation of a circle is given in standard form, you can immediately identify the center and radius by comparing it to the general equation structure.

Learn how to find the standard form of the equation of a circle with center (3, -6) and radius 6. Understand the formula and steps involved in converting the given information into the standard equation format. Suitable for students in Grades 9-12 studying geometry.

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