Question

Locus of the image of the point $(2,3)$ in the line $(2x-3y+4)+k(x-2y+3)=0$, $k\in R$ 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 radius $\sqrt{3}$

Answer 1

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

We have,

Given point are $(2,3)$

Equation of line is $(2x-3y+4)+k(x-2y+3)=0$ such that $k\in R$

From given equation,

$2x-3y+4=0......\left(1\right)$

$x-2y+3=0......\left(2\right)$

From equation (1) and (2) to and we get,

$x=1$

$y=2$

Now,

Using distance formula and we get,

Point$(2,3)$ and $(1,2)$then,

$\sqrt{{(2-1)}^2+{(3-2)}^2}$

$\sqrt{1^2+1^2}=\sqrt{2}$

Then circle of radius is $=\sqrt{2}$

Hence, this is the answer.