The parametric equations of the circle `x^(2) + y^(2) + mx + my = 0` are

Options

A. `x=-m/2+m/sqrt2cos theta , y=m/2+m/sqrt2sin theta`
B. `x=-m/2+m/sqrt2cos theta , y=-m/2+m/sqrt2sin theta`
C. x = 0 , y = 0
D. None of the above

Answer

To find the parametric equations of the circle given by the equation \(x^2 + y^2 + mx + my = 0\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
We start with the equation:
\[ x^2 + y^2 + mx + my = 0 \]
We can rearrange this to group the \(x\) and \(y\) terms:
\[ x^2 + mx + y^2 + my = 0 \]

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

For \(x\):
\[ x^2 + mx = \left(x + \frac{m}{2}\right)^2 - \frac{m^2}{4} \]

For \(y\):
\[ y^2 + my = \left(y + \frac{m}{2}\right)^2 - \frac{m^2}{4} \]

Putting these back into the equation gives:
\[ \left(x + \frac{m}{2}\right)^2 - \frac{m^2}{4} + \left(y + \frac{m}{2}\right)^2 - \frac{m^2}{4} = 0 \]

### Step 3: Simplify the Equation
Combining the terms, we have:
\[ \left(x + \frac{m}{2}\right)^2 + \left(y + \frac{m}{2}\right)^2 = \frac{m^2}{2} \]

### Step 4: Identify the Center and Radius
From the equation \(\left(x + \frac{m}{2}\right)^2 + \left(y + \frac{m}{2}\right)^2 = \frac{m^2}{2}\), we can identify:
- The center of the circle is at \(\left(-\frac{m}{2}, -\frac{m}{2}\right)\)
- The radius \(r\) is \(\sqrt{\frac{m^2}{2}} = \frac{m}{\sqrt{2}}\)

### Step 5: Write the Parametric Equations
The parametric equations for a circle centered at \((h, k)\) with radius \(r\) are given by:
\[ x = h + r \cos \theta \]
\[ y = k + r \sin \theta \]

Substituting \(h = -\frac{m}{2}\), \(k = -\frac{m}{2}\), and \(r = \frac{m}{\sqrt{2}}\):
\[ x = -\frac{m}{2} + \frac{m}{\sqrt{2}} \cos \theta \]
\[ y = -\frac{m}{2} + \frac{m}{\sqrt{2}} \sin \theta \]

### Final Parametric Equations
Thus, the parametric equations of the circle are:
\[ x = -\frac{m}{2} + \frac{m}{\sqrt{2}} \cos \theta \]
\[ y = -\frac{m}{2} + \frac{m}{\sqrt{2}} \sin \theta \]