Locus of the image of the point (2, 3) in the line `(2x-3y""+""4)""+""k""(x-2y""+""3)""=""0,""kepsiR` , is a :
(1) straight line parallel to x-axis. (2) straight line parallel to y-axis (3) circle of radius `sqrt(2)` (4) circle of radius `sqrt(3)`

Options

A. Straight line parallel to x-axis
B. straight line parallel to y-axis
C. circle of radius `sqrt(2)`
D. circle of radius 3

Answer

C

Explanation

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) -2alpha -4beta+3 = 0`
`"or " x^(2) + y^(2)-2x-4y+3=0`
which is a circle having radius,
`r = sqrt(1+4-3) = sqrt(2)`