Locus of the image of the point ( 2,3 ) in the line ( 2 x - 3 y + 4 ) + k ( x - 2 y + 3 ) = 0 , k ∈ R , is a

Let image is $( \alpha ,\beta )$(α,β)

$\therefore$∴ point $( \frac{ 2+\alpha }{2} ,\frac{ 3+\beta }{2} )$(22+α​,23+β​) lies on the given variable line.

$\therefore$∴ $2( \frac{ 2+\alpha }{2} )-3( \frac{ 3+\beta }{2} )+4+k( \frac{ 2+\alpha }{2} -2\times \frac{ 3+\beta }{2} +3 )=0$2(22+α​)−3(23+β​)+4+k(22+α​−2×23+β​+3)=0

$$\Rightarrow 2\alpha -3\beta +3+k(\alpha -2\beta +2)=0.......(1)$$⇒2α−3β+3+k(α−2β+2)=0.......(1)

again, product of two slopes $=-1$=−1

$$\Rightarrow \frac{ 3-\beta }{ 2-\alpha } \times \frac{ 2+k }{ 3+2k } =-1$$⇒2−α3−β​×3+2k2+k​=−1

$$\Rightarrow 6-2\beta +k( 3-\beta )=-( 6-3\alpha )-k( 4-2\alpha )$$⇒6−2β+k(3−β)=−(6−3α)−k(4−2α)

$$\Rightarrow k(7-2\alpha -\beta )=3\alpha +2\beta -12$$⇒k(7−2α−β)=3α+2β−12

$$\Rightarrow k=\frac{ 3\alpha +2\beta -12 }{ 7-2\alpha -\beta }$$⇒k=7−2α−β3α+2β−12​

Substituting value of $k$k in ,$(1)$(1) we get

$$2\alpha -3\beta +3+\frac{ 3\alpha +2\beta +-12 }{ 7-2\alpha -\beta } ( \alpha -2\beta +2 )=0$$2α−3β+3+7−2α−β3α+2β+−12​(α−2β+2)=0

$$\Rightarrow 14\alpha -4{\alpha }^{2} -2\alpha \beta -21\beta +6\alpha \beta +3{\beta }^{2} +21-6\alpha -3\beta +3{\alpha }^{2} -6\alpha \beta +6\alpha +2\alpha \beta -4{\beta }^{2} +4\beta -12\alpha +24\beta -24=0$$⇒14α−4α2−2αβ−21β+6αβ+3β2+21−6α−3β+3α2−6αβ+6α+2αβ−4β2+4β−12α+24β−24=0

$$\Rightarrow {\alpha }^{2} +{\beta }^{2} -2a-4\beta +3=0$$⇒α2+β2−2a−4β+3=0

$\therefore$∴ Locus is ${x}^{2} +{y}^{2} -2x-4y+3=0$x2+y2−2x−4y+3=0

Which is a circle with radius $=\sqrt{ 1+4-3 } =\sqrt{ 2 }$=1+4−3​=2​