Circle `x ^(2) + y^(2)+6y=0` touches

Options

A. Y-axis at the origin
B. X-axis at the origin
C. X-axis at the point `(3,0)`
D. Y-axis at the point `(0,2)`

Answer

To solve the problem of determining which axis the circle given by the equation \( x^2 + y^2 + 6y = 0 \) touches, we will follow these steps:

### Step 1: Rewrite the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 + 6y = 0 \]
We can rearrange it to identify the center and radius.

### Step 2: Complete the Square for the \(y\) Terms
To complete the square for the \(y\) terms, we focus on \(y^2 + 6y\):
\[ y^2 + 6y = (y + 3)^2 - 9 \]
Thus, we can rewrite the equation as:
\[ x^2 + (y + 3)^2 - 9 = 0 \]
This simplifies to:
\[ x^2 + (y + 3)^2 = 9 \]

### Step 3: Identify the Center and Radius
From the equation \(x^2 + (y + 3)^2 = 9\), we can identify:
- The center of the circle \((h, k)\) is \((0, -3)\).
- The radius \(r\) is \(\sqrt{9} = 3\).

### Step 4: Determine the Position of the Circle
The center of the circle is at \((0, -3)\) and the radius is \(3\). This means that the circle extends from:
- The lowest point at \(y = -3 - 3 = -6\)
- The highest point at \(y = -3 + 3 = 0\)

### Step 5: Analyze Touching the Axes
The circle touches the x-axis if the highest point of the circle is at \(y = 0\) (the x-axis). Since the circle reaches up to \(y = 0\) and does not go above it, it touches the x-axis.

### Conclusion
Thus, the circle touches the x-axis.