The eccentricity of the conic `x ^(2) + y^(2) -2x + 3y + 2=0` is

Options

A. 0
B. `(1)/(sqrt2)`
C. `1/2`
D. `sqrt2`

Answer

To find the eccentricity of the conic given by the equation \( x^2 + y^2 - 2x + 3y + 2 = 0 \), we will follow these steps:

### Step 1: Rearranging the equation
We start with the equation:
\[ x^2 + y^2 - 2x + 3y + 2 = 0 \]
We can rearrange this to isolate the quadratic terms:
\[ x^2 - 2x + y^2 + 3y + 2 = 0 \]

### Step 2: Completing the square
Next, we complete the square for the \(x\) and \(y\) terms.

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

For \(y^2 + 3y\):
\[ y^2 + 3y = (y + \frac{3}{2})^2 - \frac{9}{4} \]

Substituting these back into the equation gives:
\[ ((x - 1)^2 - 1) + ((y + \frac{3}{2})^2 - \frac{9}{4}) + 2 = 0 \]

### Step 3: Simplifying the equation
Now we simplify the equation:
\[ (x - 1)^2 + (y + \frac{3}{2})^2 - 1 - \frac{9}{4} + 2 = 0 \]
Combining the constants:
\[ -1 - \frac{9}{4} + 2 = -1 - 2.25 + 2 = -1.25 = -\frac{5}{4} \]
So we have:
\[ (x - 1)^2 + (y + \frac{3}{2})^2 - \frac{5}{4} = 0 \]
Rearranging gives:
\[ (x - 1)^2 + (y + \frac{3}{2})^2 = \frac{5}{4} \]

### Step 4: Identifying the conic
This equation represents a circle centered at \((1, -\frac{3}{2})\) with a radius of \(\sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}\).

### Step 5: Finding the eccentricity
For a circle, the eccentricity \(e\) is always \(0\).

Thus, the eccentricity of the conic is:
\[ \boxed{0} \]

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