A point which is inside the circle `x^(2)+y^(2)+3x-3y+2=0` is :

Options

A. (-1, 3)
B. (-2, 1)
C. (2, 1)
D. None of these

Explanation

To determine a point that lies inside the given circle defined by the equation \(x^2 + y^2 + 3x - 3y + 2 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the circle's equation in standard form. We will complete the square for both \(x\) and \(y\).

The given equation is:
\[ x^2 + y^2 + 3x - 3y + 2 = 0 \]

Rearranging gives:
\[ x^2 + 3x + y^2 - 3y + 2 = 0 \]

### Step 2: Complete the Square
For the \(x\) terms:
\[ x^2 + 3x \quad \text{can be completed as} \quad (x + \frac{3}{2})^2 - \frac{9}{4} \]

For the \(y\) terms:
\[ y^2 - 3y \quad \text{can be completed as} \quad (y - \frac{3}{2})^2 - \frac{9}{4} \]

Substituting these into the equation gives:
\[ \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} + \left(y - \frac{3}{2}\right)^2 - \frac{9}{4} + 2 = 0 \]

### Step 3: Simplify the Equation
Combining the constants:
\[ \left(x + \frac{3}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{9}{4} + 2 = 0 \]
\[ \left(x + \frac{3}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 - \frac{9}{2} + 2 = 0 \]
\[ \left(x + \frac{3}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{5}{2} \]

### Step 4: Identify the Center and Radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center: \((- \frac{3}{2}, \frac{3}{2})\)
- Radius: \(r = \sqrt{\frac{5}{2}}\)

### Step 5: Choose Points to Test
Now, we need to find points that lie inside the circle. A point \((x_1, y_1)\) lies inside the circle if:
\[ (x_1 + \frac{3}{2})^2 + (y_1 - \frac{3}{2})^2 < \frac{5}{2} \]

### Step 6: Test Points
Let's test the following points:

1. **Point A: \((-1, 3)\)**
\[ (-1 + \frac{3}{2})^2 + (3 - \frac{3}{2})^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{3}{2}\right)^2 = \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = 2.5 \]
This is equal to \(\frac{5}{2}\), so point A lies on the circle.

2. **Point B: \((-2, 1)\)**
\[ (-2 + \frac{3}{2})^2 + (1 - \frac{3}{2})^2 = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \]
This is less than \(\frac{5}{2}\), so point B lies inside the circle.

3. **Point C: \((2, 1)\)**
\[ (2 + \frac{3}{2})^2 + (1 - \frac{3}{2})^2 = \left(\frac{7}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 = \frac{49}{4} + \frac{1}{4} = \frac{50}{4} = 12.5 \]
This is greater than \(\frac{5}{2}\), so point C lies outside the circle.

### Conclusion
The only point that lies inside the circle is **Point B: \((-2, 1)\)**.