Key Concepts and Formulas

• Family of Lines: The equation $L_1 + kL_2 = 0$ represents a family of lines passing through the intersection of the lines $L_1 = 0$ and $L_2 = 0$.
• Image of a Point: The image of a point $(x_1, y_1)$ in the line $ax + by + c = 0$ is another point such that the line is the perpendicular bisector of the segment joining the point and its image.
• Locus of Image: If a family of lines passes through a fixed point A, the locus of the image of a fixed point P in these lines is a circle with center A and radius AP.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

Step-by-Step Solution

Step 1: Identify the Family of Lines and Fixed Point

The given equation represents a family of lines: $(2x - 3y + 4) + k(x - 2y + 3) = 0$. This family of lines passes through the intersection of the lines $2x - 3y + 4 = 0$ and $x - 2y + 3 = 0$. We need to find this intersection point, which will be the center of the circle representing the locus.

Step 2: Solve for the Intersection Point

Solve the following system of equations: $2x - 3y + 4 = 0 \quad (1)$ $x - 2y + 3 = 0 \quad (2)$

From equation (2), we can express $x$ in terms of $y$: $x = 2y - 3 \quad (3)$

Substitute equation (3) into equation (1): $2(2y - 3) - 3y + 4 = 0$ $4y - 6 - 3y + 4 = 0$ $y - 2 = 0$ $y = 2$

Now, substitute $y = 2$ back into equation (3): $x = 2(2) - 3$ $x = 4 - 3$ $x = 1$

Therefore, the fixed point A is $(1, 2)$.

Step 3: Identify the Fixed Point and its Image

The given fixed point is $P_0 = (2, 3)$. Let Q $(x, y)$ be the image of $P_0$ in any line from the family. Since A is a fixed point on every line in the family, $AP_0 = AQ$.

Step 4: Calculate the Radius

The radius of the circle is the distance between A $(1, 2)$ and $P_0 (2, 3)$. Using the distance formula: $AP_0 = \sqrt{(2 - 1)^2 + (3 - 2)^2}$ $AP_0 = \sqrt{1^2 + 1^2}$ $AP_0 = \sqrt{1 + 1}$ $AP_0 = \sqrt{2}$

The radius of the circle is $\sqrt{2}$.

Step 5: Determine the Equation of the Locus The locus of the image point is a circle with center $(1, 2)$ and radius $\sqrt{2}$. Therefore, the equation of the circle is: $(x - 1)^2 + (y - 2)^2 = (\sqrt{2})^2$ $(x - 1)^2 + (y - 2)^2 = 2$

Step 6: Relate the Result to the Given Options

We found that the locus is a circle with radius $\sqrt{2}$. The options are: (A) circle of radius $\sqrt{2}$ (B) circle of radius $\sqrt{3}$ (C) straight line parallel to $x$-axis (D) straight line parallel to $y$-axis

Our result matches option (A).

Common Mistakes & Tips

• Mistake: Incorrectly calculating the intersection point of the lines. Double-check the algebraic manipulations.
• Mistake: Confusing the locus of the image with the locus of the foot of the perpendicular.
• Tip: Remember that the key to this type of problem is recognizing that the locus of the image is a circle centered at the intersection of the family of lines.

Summary

The problem involves finding the locus of the image of a point in a family of lines. We identified the fixed point through which the family of lines passes by solving the system of equations. Then, we used the distance formula to calculate the radius of the circle, which is the distance between the fixed point and the given point. The locus is a circle with radius $\sqrt{2}$.

Final Answer The final answer is \boxed{\sqrt{2}}, which corresponds to option (A).