One of the diameters of the circle `x^2+y^2-12x+4y+6=0` is given by

Options

A. x+y=0
B. x+3y=0
C. x=y
D. 3x+2y=0

Answer

To solve the problem, we need to find one of the diameters of the circle given by the equation \( x^2 + y^2 - 12x + 4y + 6 = 0 \).

### Step 1: Rewrite the Circle Equation
We start by rewriting the given equation of the circle in standard form. The general form of a circle's equation is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius.

To do this, we will complete the square for both \(x\) and \(y\).

1. Rearrange the equation:
\[ x^2 - 12x + y^2 + 4y + 6 = 0 \]

2. Move the constant to the other side:
\[ x^2 - 12x + y^2 + 4y = -6 \]

3. Complete the square for \(x\):
- Take half of \(-12\) (which is \(-6\)), square it (getting \(36\)), and add it:
\[ (x^2 - 12x + 36) = (x - 6)^2 \]

4. Complete the square for \(y\):
- Take half of \(4\) (which is \(2\)), square it (getting \(4\)), and add it:
\[ (y^2 + 4y + 4) = (y + 2)^2 \]

5. Substitute back into the equation:
\[ (x - 6)^2 + (y + 2)^2 = -6 + 36 + 4 \]
\[ (x - 6)^2 + (y + 2)^2 = 34 \]

Now we have the standard form of the circle:
\[ (x - 6)^2 + (y + 2)^2 = 34 \]
This means the center of the circle is \((6, -2)\) and the radius is \(\sqrt{34}\).

### Step 2: Find the Diameter
The diameter of the circle is a line that passes through the center. We need to find which of the given options represents a line that passes through the center \((6, -2)\).

### Step 3: Check Each Option
We will substitute the point \((6, -2)\) into each of the given options to see which one satisfies the equation.

1. **Option A: \(x + y = 0\)**
\[ 6 + (-2) = 4 \quad \text{(not equal to 0, so this option is incorrect)} \]

2. **Option B: \(x + 3y = 0\)**
\[ 6 + 3(-2) = 6 - 6 = 0 \quad \text{(this option is correct)} \]

3. **Option C: \(x = y\)**
\[ 6 \neq -2 \quad \text{(not equal, so this option is incorrect)} \]

4. **Option D: \(3x + 2y = 0\)**
\[ 3(6) + 2(-2) = 18 - 4 = 14 \quad \text{(not equal to 0, so this option is incorrect)} \]

### Conclusion
The only option that satisfies the condition of passing through the center of the circle is **Option B: \(x + 3y = 0\)**.

### Final Answer
**One of the diameters of the circle is given by: \(x + 3y = 0\)**.

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