Question

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

• A. Straight line parallel to x-axis
• B. Straight line parallel to y-axis
• C. Circle of radius $\sqrt{2}$
• D. Circle of adius 3

Answer 1

The correct option is C Circle of radius $\sqrt{2}$

Lines $(2x-3y+4)+k(x-2y+3)=0$ pass throught point of intersection of $2x-3y+4=0$ and $x-2y+3=0$ which is $A(1,2)$

Let image of P (2,3) in any line throught A be $Q(\alpha ,\beta ).$
$\therefore AP=AQ\therefore (\alpha -1{)}^2+(\beta -2{)}^2=(2-1{)}^2+(3-2{)}^2\therefore {\alpha }^2+{\beta }^2-2\alpha -4\beta +3=0or{x}^2+y^2-2x-4y+3=0,$
which is a circle having radius,
$r=\sqrt{1+4-3}=\sqrt{2}$