Proving that the locus of the image of $(2, 3)$ in the line $2x-3y+4+k(x-2y+3) =0$ is a circle of radius $\sqrt2$.

Question

Prove that the locus of the image of the point $(2, 3)$ in the line $2x-3y+4+k(x-2y+3) =0$, for $k$ real, is a circle of radius $\sqrt2$.

My approach was finding the equation of the line perpendicular to the given one and passing through (2, 3). After that finding out the point of intersection proved to be humongous for me. And that is where I got stuck.

Answer 1

Let $Q(2,3)$

$P(h,k)$ be the required image

$M$ be the midpoint of $PQ$ i.e., $\left(\dfrac{h+2}2,\dfrac{k+3}2\right)$

The intersection of $2x-3y+4=0; x-2y+3=0$ is $R(1,2)$

Now $MR\perp PQ$

$$\implies\dfrac{k-3}{h-2}\cdot\dfrac{\dfrac{k+3}2-2}{\dfrac{h+2}2-1}=-1$$

$$\iff(h-1)^2+(k-2)^2=2$$

Answer 2

For $k \in \Bbb{R}$, the line $L_1+kL_2=0$, represents the family of lines that pass through the intersection of the lines $L_1=0$ and $L_2=0$. The intersection point is $C(1,2)$.

In the picture below, let the given line be an arbitrary line that passes through the point $C(1,2)$. We reflect the given point $D(2,3)$ with respect to this line. Let $E(s,t)$ be the reflection of $D$. Then $DC=EC$. This means $$(s-1)^2+(t-2)^2=(3-2)^2+(2-1)^2$$

So locus would be the circle with center at $(1,2)$ and radius equal to the distance between $(1,2)$ and $(2,3)$

Answer 3

P = $(2,3)$ Line L:   $(2x - 3y + 4) + k (x - 2y +3) = 0 $   $k ∈ R$

L passes through intersection point of  $2x-3y +4 =0$ and $x-2y+3 = 0$

Calculating point:  C$(1,2)$.

image points Q $(α, β)$ of P$(2,3$) wrt line L.

Shift Origin from O $(0,0)$ to C $(1,2)$ with new coordinate system of $X = x-1$ and $Y= y-2.$ We need to find the images of P$(X=2-1=1, Y=3-2=1)$ on lines passing thru C$(X=0, Y=0)$.

Clearly we can see that points $(X,Y)=(-1,-1), (-1, 1), (1,1), (1,-1)$ are image points wrt $X=-Y, X=0, X= Y, Y= 0$   respectively. Clearly the locus is a circle of radius = $CP = √(1²+1²) = √2$ Locus:   $$X² + Y² = 2$$  $$(x-1)² + (y-2)² = 2$$