The eccentricity of the hyperbola `|sqrt((x-3)^2+(y-2)^2)-sqrt((x+1)^2+(y+1)^2)|=1` is ______

Explanation

To find the eccentricity of the hyperbola given by the equation

\[ |\sqrt{(x-3)^2 + (y-2)^2} - \sqrt{(x+1)^2 + (y+1)^2}| = 1, \]

we will follow these steps:

### Step 1: Identify the foci
The equation of the hyperbola can be interpreted as the difference in distances from any point \( P(x,y) \) to the two fixed points (foci) \( S(3, 2) \) and \( S'(-1, -1) \).

### Step 2: Calculate the distance between the foci
The distance \( d \) between the foci \( S \) and \( S' \) can be calculated using the distance formula:

\[ d = \sqrt{(3 - (-1))^2 + (2 - (-1))^2} = \sqrt{(3 + 1)^2 + (2 + 1)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5. \]

### Step 3: Relate the distance to the hyperbola parameters
For a hyperbola, the distance between the foci \( 2c \) is given by:

\[ 2c = d. \]

From our calculation, \( d = 5 \), so:

\[ 2c = 5 \implies c = \frac{5}{2} = 2.5. \]

### Step 4: Use the property of hyperbola
For a hyperbola, the relationship between the semi-major axis \( a \), semi-minor axis \( b \), and the distance to the foci \( c \) is given by:

\[ c^2 = a^2 + b^2. \]

### Step 5: Find the value of \( a \)
From the hyperbola's property, we know that the difference in distances (the right side of the original equation) is \( 2a = 1 \). Thus:

\[ a = \frac{1}{2}. \]

### Step 6: Calculate \( b \)
Now we can find \( b \) using the relationship \( c^2 = a^2 + b^2 \):

\[ c^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4}, \]
\[ a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}. \]

Substituting these values into the equation:

\[ \frac{25}{4} = \frac{1}{4} + b^2. \]

Solving for \( b^2 \):

\[ b^2 = \frac{25}{4} - \frac{1}{4} = \frac{24}{4} = 6. \]

### Step 7: Find the eccentricity \( e \)
The eccentricity \( e \) of a hyperbola is given by:

\[ e = \frac{c}{a}. \]

Substituting the values we have:

\[ e = \frac{c}{a} = \frac{\frac{5}{2}}{\frac{1}{2}} = 5. \]

### Final Answer
Thus, the eccentricity of the hyperbola is

\[ \boxed{5}. \]