Locus of image of point in a line. I am given the following question:

Find the locus of the image of the point $(2,3)$ in the line
$$\text{L}:(2x-3y+4)+k(x-2y+3)=0$$ where $k$ is any real number.

Attempt at solution.
I used a formula to arrive at (image of a point):
$\dfrac{x-2}{(2+k)}=\dfrac{y-3}{-(2k+3)}=\dfrac{2(1+k)}{(k+2)^2+(2k+3)^2}$
I know that it is a circle centred at $(1,2)$ with radius $\sqrt{2}$.
But I am having trouble solving the parametric equation. I need a quick way to do it or maybe to solve this problem in other way. It appeared in a recent National exam in India known as JEE - Mains 2015.
Addendum : Here's a graph https://www.desmos.com/calculator/5bweaiyj3a
 A: 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 
$$(a-1)^2+(b-2)^2=2$$
A: 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 $\sqrt2$
