The parametric equations of the circle
`x^(2)+y^(2)+x+sqrt(3)y=0` are ............

Answer

To find the parametric equations of the circle given by the equation \(x^2 + y^2 + x + \sqrt{3}y = 0\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
We start with the given equation:
\[ x^2 + y^2 + x + \sqrt{3}y = 0 \]
We will rearrange this equation to group the \(x\) and \(y\) terms.

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

1. For \(x\):
\[ x^2 + x = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} \]

2. For \(y\):
\[ y^2 + \sqrt{3}y = \left(y + \frac{\sqrt{3}}{2}\right)^2 - \frac{3}{4} \]

Now substituting back into the equation:
\[ \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + \left(y + \frac{\sqrt{3}}{2}\right)^2 - \frac{3}{4} = 0 \]
This simplifies to:
\[ \left(x + \frac{1}{2}\right)^2 + \left(y + \frac{\sqrt{3}}{2}\right)^2 = 1 \]

### Step 3: Identify the Center and Radius
From the standard form of the circle \((x - a)^2 + (y - b)^2 = r^2\), we can identify:
- Center \((a, b) = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)\)
- Radius \(r = 1\)

### Step 4: Write the Parametric Equations
The parametric equations for a circle are given by:
\[ x = a + r \cos \theta \]
\[ y = b + r \sin \theta \]
Substituting the values of \(a\), \(b\), and \(r\):
\[ x = -\frac{1}{2} + 1 \cdot \cos \theta = -\frac{1}{2} + \cos \theta \]
\[ y = -\frac{\sqrt{3}}{2} + 1 \cdot \sin \theta = -\frac{\sqrt{3}}{2} + \sin \theta \]

### Final Parametric Equations
Thus, the parametric equations of the circle are:
\[ x = -\frac{1}{2} + \cos \theta \]
\[ y = -\frac{\sqrt{3}}{2} + \sin \theta \]

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