Locus of image of point in a line.

The given line is actually a variable line (or you can say a family of lines) that pass through the intersection of the following two lines: $$2x-3y+4=0 \qquad \text{ and } \qquad x-2y+3=0.$$ Their intersection point is $J=(1,2)$.

Let $P=(a,b)$ be the reflection of $Q=(2,3)$, then the mid-point $M=\left(\dfrac{a+2}{2}, \dfrac{b+3}{2}\right)$ of segment $PQ$ lies on the given line.

Now use a bit of geometric imagination to see that the lines $JM$ and $PQ$ will be perpendicular to each other (except in the degenerate case which happens when the variable line passes through the point $(2,3)$ itself). Thus the slopes will follow: $$m_{JM} \cdot m_{PQ}=-1$$ This is same as saying $$\left(\frac{\frac{b+3}{2}-2}{\frac{a+2}{2}-1}\right).\left(\frac{b-3}{a-2}\right)=-1.$$ As you can see there is no parameter in this equation. Once you solve this you get: $$a^2+b^2-2a-4b+3=0$$ which is same as $$(a-1)^2+(b-2)^2=2$$

After getting the point of intersection of the given family of lines we need to simply calculate the distance between the given point $(2,3)$ and the point of intersection of the family of lines i.e., $(1,2)$ which is equal to $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ so $$d= \sqrt{(2-1)^2+(3-2)^2}=\sqrt2$$ Now by the definition of locus of a point (locus of a point at a distance d from a given point is a circle of radius $d$) we can say that the locus of the required point is a circle of radius $\sqrt2$