Question

Let $A\left(2,3\right),B\left(4,5\right)$ be two points and let $C\equiv \left(x,y\right)$ be a point such that $\left(x-2\right)\left(x-4\right)+\left(y-3\right)\left(y-5\right)=0$ . If area of $\Delta ABC=\sqrt{2}$ $sq.unit$ , then maximum number of positions of $C$ in the $xy$ plane is :

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (D)

We have,
$\left(x-2\right)\left(x-4\right)+\left(y-3\right)\left(y-5\right)=0$
$\Rightarrow x^2+y^2-6x-8y+23=0$
$\Rightarrow x^2+y^2-6x-8y+16+9=2$
$\Rightarrow {\left(x-3\right)}^2+{\left(y-4\right)}^2={\left(\sqrt{2}\right)}^2$
This is the equation of circle with centre $\left(3,4\right)$ and radius $\sqrt{2}$ units.

Now, $AB=2\sqrt{2}$
$\text{ar}\left(\Delta ABC\right)=\frac{1}{2}\left(AB\right)h$
$\Rightarrow \sqrt{2}=\frac{1}{2}\left(2\sqrt{2}\right)h$
$h=1$
$\therefore$ There are maximum four positions.