Locus of the image of the point (2, 3) in the line `(2x-3y""+""4)""+""k""(x-2y""+""3)""=""0,""kepsiR` , is a :
(1) straight line parallel to x-axis. (2) straight line parallel to y-axis (3) circle of radius `sqrt(2)` (4) circle of radius `sqrt(3)`

Options

A. Straight line parallel to x-axis
B. straight line parallel to y-axis
C. circle of radius `sqrt(2)`
D. circle of radius 3

Explanation

To find the locus of the image of the point (2, 3) in the line given by the equation \( (2x - 3y + 4) + k(x - 2y + 3) = 0 \), we will follow these steps:

### Step 1: Understand the given family of lines
The equation can be rewritten as:
\[ 2x - 3y + 4 + k(x - 2y + 3) = 0 \]
This represents a family of lines parameterized by \( k \).

### Step 2: Identify the two lines
We can separate the equation into two parts:
1. \( 2x - 3y + 4 = 0 \) (Line 1)
2. \( x - 2y + 3 = 0 \) (Line 2)

### Step 3: Solve for the intersection of the two lines
To find the intersection point of these two lines, we can solve them simultaneously.

From Line 2, we can express \( x \) in terms of \( y \):
\[ x = 2y - 3 \]

Now substitute this expression for \( x \) into Line 1:
\[ 2(2y - 3) - 3y + 4 = 0 \]
Expanding this gives:
\[ 4y - 6 - 3y + 4 = 0 \]
Combining like terms:
\[ y - 2 = 0 \]
Thus, we find:
\[ y = 2 \]

Now substitute \( y = 2 \) back into the expression for \( x \):
\[ x = 2(2) - 3 = 4 - 3 = 1 \]
So the intersection point is \( (1, 2) \).

### Step 4: Find the distance from the point (2, 3) to the intersection point (1, 2)
Let \( A = (2, 3) \) and \( B = (1, 2) \). The distance \( AB \) can be calculated using the distance formula:
\[ AB = \sqrt{(2 - 1)^2 + (3 - 2)^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \]

### Step 5: Set up the equation for the locus of the image
Let the image of point \( A \) be \( (h, k) \). The distance from \( A \) to \( B \) must equal the distance from \( B \) to the image point \( (h, k) \):
\[ AB = AP \]
Thus, we have:
\[ \sqrt{(2 - 1)^2 + (3 - 2)^2} = \sqrt{(h - 1)^2 + (k - 2)^2} \]
Squaring both sides gives:
\[ 2 = (h - 1)^2 + (k - 2)^2 \]

### Step 6: Replace \( h \) and \( k \) with \( x \) and \( y \)
For the locus, we replace \( h \) with \( x \) and \( k \) with \( y \):
\[ (x - 1)^2 + (y - 2)^2 = 2 \]

### Conclusion: Identify the locus
The equation \( (x - 1)^2 + (y - 2)^2 = 2 \) represents a circle with:
- Center: \( (1, 2) \)
- Radius: \( \sqrt{2} \)

Thus, the locus of the image of the point \( (2, 3) \) is a circle of radius \( \sqrt{2} \).

### Final Answer:
The correct option is (3) circle of radius \( \sqrt{2} \).