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.