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The given equation is:

\[
x^2 + y^2 - 2x - 6y - 26 = 0
\]

To write this in standard form, we need to complete the square for both \( x \) and \( y \).

### Step 1: Group the \( x \) and \( y \) terms
\[
(x^2 - 2x) + (y^2 - 6y) = 26
\]

### Step 2: Complete the square
For \( x^2 - 2x \), take half of the coefficient of \( x \), square it, and add it:
\[
\left( \frac{-2}{2} \right)^2 = 1
\]
Add and subtract 1 inside the \( x \)-term.

For \( y^2 - 6y \), take half of the coefficient of \( y \), square it, and add it:
\[
\left( \frac{-6}{2} \right)^2 = 9
\]
Add and subtract 9 inside the \( y \)-term.

So the equation becomes:
\[
(x^2 - 2x + 1 - 1) + (y^2 - 6y + 9 - 9) = 26
\]
\[
(x - 1)^2 - 1 + (y - 3)^2 - 9 = 26
\]

### Step 3: Simplify
\[
(x - 1)^2 + (y - 3)^2 - 10 = 26
\]
Move the constant \( -10 \) to the right-hand side:
\[
(x - 1)^2 + (y - 3)^2 = 36
\]

### Step 4: Write in standard form
The standard form of a circle is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
From this, we can see that the center of the circle is \( (1, 3) \) and the radius is:
\[
r = \sqrt{36} = 6
\]

### Final Answer:
- **Standard Form:** \( (x - 1)^2 + (y - 3)^2 = 36 \)
- **Center:** \( (1, 3) \)
- **Radius:** 6

Would you like more details on the graphing part or have any further questions?

Here are some additional related questions for you to explore:
1. How do you graph a circle from its standard equation?
2. What is the relationship between the general form and the standard form of a circle?
3. How would you complete the square if the coefficients of \( x^2 \) and \( y^2 \) were not 1?
4. What changes if the constant on the right-hand side is negative?
5. How do transformations affect the standard equation of a circle?

**Tip:** When completing the square, always remember to balance the equation by adding or subtracting the same value on both sides!

Complete the square and write the given equation in standard form. Then give the center and radius of the circle and graph the equation: x^2 + y^2 - 2x - 6y - 26 = 0.

Learn how to complete the square and write the equation x^2 + y^2 - 2x - 6y - 26 = 0 in standard form. This step-by-step guide also provides the center and radius of the resulting circle, suitable for Grades 9-12.

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