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 if a point is inside the 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
We start with the equation of the circle:
\[ x^2 + y^2 + 3x - 3y + 2 = 0 \]
We can rearrange this into the standard form of a circle by completing the square.

### Step 2: Complete the Square
1. For \(x^2 + 3x\):
- Take half of the coefficient of \(x\) (which is \(\frac{3}{2}\)), square it to get \(\left(\frac{3}{2}\right)^2 = \frac{9}{4}\).
- Rewrite:
\[ x^2 + 3x = \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} \]

2. For \(y^2 - 3y\):
- Take half of the coefficient of \(y\) (which is \(-\frac{3}{2}\)), square it to get \(\left(-\frac{3}{2}\right)^2 = \frac{9}{4}\).
- Rewrite:
\[ y^2 - 3y = \left(y - \frac{3}{2}\right)^2 - \frac{9}{4} \]

### Step 3: Substitute Back into the Equation
Substituting these back into the original equation gives:
\[ \left(x + \frac{3}{2}\right)^2 - \frac{9}{4} + \left(y - \frac{3}{2}\right)^2 - \frac{9}{4} + 2 = 0 \]
Simplifying this:
\[ \left(x + \frac{3}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 - \frac{9}{2} + 2 = 0 \]
\[ \left(x + \frac{3}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{5}{2} \]

### Step 4: Identify the Center and Radius
From the equation \(\left(x + \frac{3}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{5}{2}\):
- The center of the circle is \((- \frac{3}{2}, \frac{3}{2})\).
- The radius \(r\) is \(\sqrt{\frac{5}{2}}\).

### Step 5: Check Points for Inside the Circle
To check if a point \((x_1, y_1)\) is inside the circle, we calculate the distance from the center to the point and compare it to the radius.

#### Example Point A: \((-1, 3)\)
1. Calculate the distance \(OA\):
\[ OA = \sqrt{\left(-1 + \frac{3}{2}\right)^2 + \left(3 - \frac{3}{2}\right)^2} \]
\[ = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{3}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{9}{4}} = \sqrt{\frac{10}{4}} = \sqrt{\frac{5}{2}} \]
Since \(OA = r\), point A lies on the circle.

#### Example Point B: \((-2, 1)\)
1. Calculate the distance \(OB\):
\[ OB = \sqrt{\left(-2 + \frac{3}{2}\right)^2 + \left(1 - \frac{3}{2}\right)^2} \]
\[ = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} \]
Since \(OB < r\), point B lies inside the circle.

### Conclusion
The point \((-2, 1)\) is inside the circle.