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To solve the problem, we need to find the center and radius of the given circle from its general equation:

\[
x^2 - 6x + y^2 + 12y + 20 = 0
\]

### Step 1: Rewrite the equation in standard form
The standard form of a circle's equation is:

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

Where \((h, k)\) is the center, and \(r\) is the radius. We can convert the given equation into this form by completing the square for both \(x\) and \(y\).

### Step 2: Completing the square for \(x\)
Take the \(x\)-terms: \(x^2 - 6x\).

To complete the square, take half of the coefficient of \(x\) (which is \(-6\)), square it, and add it inside the equation. So:
\[
\left(\frac{-6}{2}\right)^2 = (-3)^2 = 9
\]

Add 9 inside the equation and subtract 9 outside to keep the equation balanced.

\[
x^2 - 6x + 9 = (x - 3)^2
\]

### Step 3: Completing the square for \(y\)
Now take the \(y\)-terms: \(y^2 + 12y\).

Take half of the coefficient of \(y\) (which is \(12\)), square it, and add it inside the equation. So:
\[
\left(\frac{12}{2}\right)^2 = 6^2 = 36
\]

Add 36 inside the equation and subtract 36 outside to keep the equation balanced.

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

### Step 4: Rewrite the equation
Now the equation looks like this:
\[
(x - 3)^2 + (y + 6)^2 = 9 + 36 - 20
\]
\[
(x - 3)^2 + (y + 6)^2 = 25
\]

### Step 5: Identify the center and radius
Now we can compare this with the standard form \((x - h)^2 + (y - k)^2 = r^2\).

- The center \((h, k)\) is \((3, -6)\).
- The radius \(r\) is \(\sqrt{25} = 5\).

### Final answer:
- **x-value of the center**: 3
- **y-value of the center**: -6
- **Radius of the circle**: 5

Would you like more details, or do you have any questions? 

Here are 5 related questions to deepen your understanding:
1. How do you complete the square for any quadratic equation?
2. What happens if the constant term in the circle equation is negative?
3. How do you derive the general form of a circle from its standard form?
4. Can a circle equation ever have a negative radius? Why or why not?
5. How do you find the equation of a circle passing through a given point?

**Tip**: Always remember to balance both sides of the equation when completing the square.

We wish to find the properties of the circle x^2 - 6x + y^2 + 12y + 20 = 0. Find the center and the radius of the circle.

Learn how to find the center and radius of the circle given by the equation x^2 - 6x + y^2 + 12y + 20 = 0. This problem demonstrates the method of completing the square for both x and y terms and converting the equation into standard form.

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