Describe the circles x²+y²-6x+4y=0 and x²+y²-6x+4y-52=0

Let's analyze the given equations of the circles.

### Circle 1: \( x^2 + y^2 - 6x + 4y = 0 \)

1. **Standard Form**: To describe the circle, we need to rewrite the equation in the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center, and \(r\) is the radius.
   
2. **Completing the Square**:
   - Group the \(x\) and \(y\) terms:  
     \[
     (x^2 - 6x) + (y^2 + 4y) = 0
     \]
   - Complete the square for \(x\):
     \[
     x^2 - 6x = (x - 3)^2 - 9
     \]
   - Complete the square for \(y\):
     \[
     y^2 + 4y = (y + 2)^2 - 4
     \]
   - Substitute these into the equation:
     \[
     (x - 3)^2 - 9 + (y + 2)^2 - 4 = 0
     \]
   - Simplify:
     \[
     (x - 3

Explore the analysis of circle equations x²+y²-6x+4y=0 and x²+y²-6x+4y-52=0. Understand the standard form of a circle, completing the square method, and geometric interpretations. Suitable for advanced high school students.

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