The given equation is:

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

This equation represents a circle in the Cartesian plane. To rewrite it in the standard form of a circle's equation, we need to complete the square for the \( y \) term.

1. Start by isolating the \( y \)-related terms:
   \[
   x^2 + (y^2 - 6y) = 0
   \]
   
2. To complete the square for the \( y \) term:
   \[
   y^2 - 6y
   \]
   Take half of the coefficient of \( y \) (which is -6), square it, and add it inside the parenthesis. 
   \[
   \left(\frac{-6}{2}\right)^2 = 9
   \]
   
   Add 9 inside the equation, but since you are adding it to the left side, you must also subtract it from the right side:
   \[
   x^2 + (y^2 - 6y + 9) = 9
   \]
   
3. The equation now simplifies to:
   \[
   x^2 + (y - 3)^2 = 9
   \]

This is the equation of a circle with a center at \((0, 3)\) and a radius of \(3\) (since \( \sqrt{9} = 3 \)).

**Summary:**
- **Center**: \((0, 3)\)
- **Radius**: \(3\)

Would you like more details or have any questions?

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Here are some related questions:

1. What is the general form of a circle's equation?
2. How do you complete the square for a general quadratic equation?
3. Can you explain how the center and radius of a circle are derived from its equation?
4. How would the equation change if the circle were translated in the x-direction?
5. How would the equation look if the radius were doubled?
6. What is the geometric significance of completing the square in this context?
7. How can we convert a general conic equation into a standard form?
8. What are the properties of a circle derived from its equation?

**Tip:** Completing the square is a useful technique not only in geometry but also in solving quadratic equations and optimization problems.

Learn how to solve the circle equation x² + y² - 6y = 0 by completing the square. This problem demonstrates the application of completing the square technique in algebra, specifically in circle equations. Suitable for students in Grades 10-12.

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