The equation of a diameter of circle `x ^(2) + y^(2)-6x + 2y =0`, passing through origin is

Options

A. `x+3y=0`
B. `x-3y =0`
C. `3x+ y=0`
D. `3x-y=0`

Answer

To find the equation of a diameter of the circle given by the equation \(x^2 + y^2 - 6x + 2y = 0\) that passes through the origin, we can follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the circle's equation in standard form. The given equation is:
\[ x^2 + y^2 - 6x + 2y = 0 \]
We can rearrange it to:
\[ x^2 - 6x + y^2 + 2y = 0 \]

### Step 2: Complete the Square
Next, we will complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 - 6x \rightarrow (x - 3)^2 - 9 \]

For \(y\):
\[ y^2 + 2y \rightarrow (y + 1)^2 - 1 \]

Now substituting these back into the equation:
\[ (x - 3)^2 - 9 + (y + 1)^2 - 1 = 0 \]
This simplifies to:
\[ (x - 3)^2 + (y + 1)^2 - 10 = 0 \]
or
\[ (x - 3)^2 + (y + 1)^2 = 10 \]

### Step 3: Identify the Center and Radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \((h, k) = (3, -1)\)
- Radius \(r = \sqrt{10}\)

### Step 4: Find the Equation of the Diameter
The diameter passes through the center \((3, -1)\) and the origin \((0, 0)\). We can use the two points to find the equation of the line (diameter).

Using the two-point form of the equation of a line:
\[ \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} \]
Substituting \((x_1, y_1) = (3, -1)\) and \((x_2, y_2) = (0, 0)\):
\[ \frac{y - (-1)}{0 - (-1)} = \frac{x - 3}{0 - 3} \]
This simplifies to:
\[ \frac{y + 1}{1} = \frac{x - 3}{-3} \]
Cross-multiplying gives:
\[ y + 1 = -\frac{1}{3}(x - 3) \]
Multiplying through by 3 to eliminate the fraction:
\[ 3(y + 1) = -(x - 3) \]
Expanding:
\[ 3y + 3 = -x + 3 \]
Rearranging gives:
\[ x + 3y = 0 \]

### Final Equation
Thus, the equation of the diameter of the circle passing through the origin is:
\[ x + 3y = 0 \]