Locus of the image of the point (2, 3) in the line $$(2x - 3y + 4) + k(x - 2y + 3) = 0$$, k $$\in \mathbb{R}$$, is a

The given equation is $$(2x - 3y + 4) + k(x - 2y + 3) = 0$$. This represents a family of lines passing through the intersection of:

$$L_1: 2x - 3y + 4 = 0$$ and $$L_2: x - 2y + 3 = 0$$

To find the intersection point ($$P$$), solve the two equations

$$\implies \mathbf{y = 2}$$, $$\mathbf{x = 1}$$.

Thus, all lines in the family pass through the fixed point $$P(1, 2)$$.

Let $$A(2, 3)$$ be the given point and $$A'$$ be its image in any line $$L$$ from the family.

By the definition of a reflection, the line $$L$$ is the perpendicular bisector of the segment $$AA'$$.

Since the fixed point $$P(1, 2)$$ lies on every such line $$L$$, it must be equidistant from the point $$A$$ and its image $$A'$$.

$$\text{Therefore, } PA = PA'$$

$$PA = \sqrt{(2-1)^2 + (3-2)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$$

Since $$PA' = PA = \sqrt{2}$$, the image $$A'$$ always remains at a constant distance of $$\sqrt{2}$$ from the fixed point $$P(1, 2)$$.

The locus of a point moving at a constant distance from a fixed point is a circle, 

with Center: $$(1, 2)$$, and Radius: $$\sqrt{2}$$