Given family of lines

The line is $(2x-3y+4)+k(x-2y+3)=0, \quad k\in \mathbb R.$

Rearranging, $(2+k)x + (-3-2k)y + (4+3k)=0.$

Let this line be $ax+by+c=0$ where $a=2+k,\quad b=-3-2k,\quad c=4+3k.$

Image of point $(2,3)$ in the line

If $(x_1,y_1)$ is reflected in the line $ax+by+c=0$, then the image $(x',y')$ is $x' = x_1 - \frac{2a(ax_1+by_1+c)}{a^2+b^2},$ $y' = y_1 - \frac{2b(ax_1+by_1+c)}{a^2+b^2}.$

Here $(x_1,y_1)=(2,3)$.

Compute $ax_1+by_1+c$

$a(2)+b(3)+c = 2(2+k)+3(-3-2k)+(4+3k).$

Simplify: $= (4+2k)+(-9-6k)+(4+3k)$ $= -1-k.$

Also, $a^2+b^2=(2+k)^2+(-3-2k)^2.$

Compute: $(2+k)^2 = k^2+4k+4,$ $(-3-2k)^2 = 4k^2+12k+9.$

Hence, $a^2+b^2 = 5k^2+16k+13.$

Coordinates of reflected point

Therefore, $x = 2 - \frac{2(2+k)(-1-k)}{5k^2+16k+13}$ $= 2 + \frac{2(2+k)(1+k)}{5k^2+16k+13},$

$y = 3 - \frac{2(-3-2k)(-1-k)}{5k^2+16k+13}$ $= 3 - \frac{2(3+2k)(1+k)}{5k^2+16k+13}.$

Use geometric property of the family

A simpler approach is to find the fixed point of the family of lines.

Since $(2x-3y+4)+k(x-2y+3)=0$ holds for all $k$, the common point must satisfy $2x-3y+4=0$ and $x-2y+3=0.$

Solve these:

From $x=2y-3.$ Substitute into the first: $2(2y-3)-3y+4=0$ $4y-6-3y+4=0$ $y-2=0 \Rightarrow y=2.$ Then $x=2(2)-3=1.$

So every line in the family passes through the fixed point $(1,2)$.

Key reflection fact

The image of a fixed point $P=(2,3)$ in any line through a fixed point $A=(1,2)$ lies on a circle centered at $A$ with radius $AP$.

Why? Reflection preserves distance from the mirror line, and since $A$ lies on the reflecting line, its distance from $P$ equals its distance from the image $P'$. Thus $AP' = AP.$

Now, $AP = \sqrt{(2-1)^2+(3-2)^2} = \sqrt{1+1} = \sqrt{2}.$

Hence the locus is the circle $(x-1)^2+(y-2)^2=2.$

Match with options

This is a circle of radius $\sqrt{2}$.

Therefore, the correct option is: $\boxed{\text{A}}$