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  4. Let A(2 , 3),B(4 , 5) be two points and let C≡ (x , y) be a point such that (x - 2)(x - 4)+(y - 3)(y - 5)=0 . If area of Δ ABC=√2 sq.unit , then maximum number of positions of C in the xy plane is :

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Q. 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 :

NTA AbhyasNTA Abhyas 2022

Solution:

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