Standard form of a circle state center and radius. X^2+y^2-6y-7=0

To rewrite the equation \( x^2 + y^2 - 6y - 7 = 0 \) in standard form, we'll complete the square for the \(y\)-terms.

### Step-by-step solution:
1. Start with the given equation:
   \[
   x^2 + y^2 - 6y - 7 = 0
   \]
   
2. Move the constant term to the other side:
   \[
   x^2 + y^2 - 6y = 7
   \]

3. Complete the square for the \( y \)-terms:
   - The \( y^2 - 6y \) part can be rewritten by adding and subtracting \(\left(\frac{-6}{2}\right)^2 = 9\):
     \[
     y^2 - 6y + 9 = (y - 3)^2
     \]
   
4. Add \( 9 \) to both sides of the equation to keep it balanced:
   \[
   x^2 + (y - 3)^2 = 7 + 9
   \]
   \[
   x^2 + (y - 3)^2 = 16
   \]

### Standard form of the circle:
The equation is now in standard form:
\[
(x - 0)^2 + (y - 3)^2 = 4^2
\]

### Center and Radius:
- **Center**: \( (0, 3) \)
- **Radius**: \( 4 \)

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Do you want more details on this process or have any questions? Here are 5 related questions to explore further:

1. How do you complete the square for more complex expressions?
2. How do you find the equation of a circle given two points on the circle?
3. What is the general form of the equation of a circle?
4. How do you derive the equation of a circle from its geometric definition?
5. Can you convert the equation of a circle back to its expanded form?

**Tip:** When completing the square, always ensure you balance the equation by adding the same value to both sides.

Standard form of a circle state center and radius. X^2 + y^2 - 6y - 7 = 0

Learn how to convert the equation X^2 + Y^2 - 6Y - 7 = 0 into the standard form of a circle using the method of completing the square. This solution explains how to identify the center and radius of the circle, ideal for students in Grades 9-12.

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