Equation of the diameter of the circle `x^(2) +y^(2) - 2x + 4y` = 0 which passes through the origin is

Options

A. `x + 2y = 0 `
B. `x - 2y = 0 `
C. `2x + y = 0 `
D. `2x - y = 0 `

Answer

To find the equation of the diameter of the circle \(x^2 + y^2 - 2x + 4y = 0\) that passes through the origin, we can follow these steps:

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ x^2 + y^2 - 2x + 4y = 0 \]
We can rearrange this equation to identify the center and radius of the circle.

### Step 2: Complete the square for \(x\) and \(y\)
To complete the square for \(x\) and \(y\):

1. For \(x^2 - 2x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

2. For \(y^2 + 4y\):
\[ y^2 + 4y = (y + 2)^2 - 4 \]

### Step 3: Substitute back into the equation
Substituting these completed squares back into the equation gives:
\[ (x - 1)^2 - 1 + (y + 2)^2 - 4 = 0 \]
This simplifies to:
\[ (x - 1)^2 + (y + 2)^2 - 5 = 0 \]
Thus, we can rewrite it as:
\[ (x - 1)^2 + (y + 2)^2 = 5 \]
This indicates that the center of the circle is at \((1, -2)\) and the radius is \(\sqrt{5}\).

### Step 4: Find the slope of the diameter
The diameter of the circle that passes through the origin \((0, 0)\) and the center \((1, -2)\) can be found by calculating the slope of the line connecting these two points:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 0}{1 - 0} = \frac{-2}{1} = -2 \]

### Step 5: Write the equation of the line
Using the point-slope form of the equation of a line, which is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point through which the line passes (in this case, the origin \((0, 0)\)):
\[ y - 0 = -2(x - 0) \]
This simplifies to:
\[ y = -2x \]

### Conclusion
The equation of the diameter of the circle that passes through the origin is:
\[ y = -2x \]