To convert the circle defined by center \( C(2, 6) \) and radius \( r = 3 \) into its general equation, we follow 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:
- \( h = 2 \),
- \( k = 6 \),
- \( r = 3 \).

Substitute these values into the equation:

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

Simplifying:

\[
(x - 2)^2 + (y - 6)^2 = 9
\]

This is the equation of the circle in its standard form.

If you want the **general form** of the equation, expand the terms:

\[
(x - 2)^2 = x^2 - 4x + 4
\]
\[
(y - 6)^2 = y^2 - 12y + 36
\]

Now, combine everything:

\[
x^2 - 4x + 4 + y^2 - 12y + 36 = 9
\]

Simplifying further:

\[
x^2 + y^2 - 4x - 12y + 40 = 9
\]

Finally, subtract 9 from both sides:

\[
x^2 + y^2 - 4x - 12y + 31 = 0
\]

This is the **general equation** of the circle.

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Would you like more details or have any questions?

Here are 5 related questions you might consider:
1. How do you find the center of a circle given its general equation?
2. How would the equation change if the radius was doubled?
3. What happens to the equation if the center is moved to the origin?
4. How do you find the points of intersection between two circles?
5. How do you rewrite a general equation in standard form?

**Tip:** Always ensure to subtract or add constants carefully when converting between forms of equations.

Learn how to convert the equation of a circle centered at (2, 6) with a radius of 3 into its general form. This problem covers key geometric concepts including the equation of a circle, standard form, and conversion to general form. Suitable for advanced high school students.

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