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!