Let A(2, 3), B(4, 5) be two points and let C | vedclass

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(D) The given equation $(x - 2)(x - 4) + (y - 3)(y - 5) = 0$ represents a circle with diameter $AB$,where $A = (2, 3)$ and $B = (4, 5)$.The length of

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$4$

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(D) The given equation $(x - 2)(x - 4) + (y - 3)(y - 5) = 0$ represents a circle with diameter $AB$,where $A = (2, 3)$ and $B = (4, 5)$.The length of the diameter $AB = \sqrt{(4 - 2)^2 + (5 - 3)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes h = \sqrt{2}$.Substituting $AB = 2\sqrt{2}$,we get $\frac{1}{2} 	imes 2\sqrt{2} 	imes h = \sqrt{2}$,which simplifies to $\sqrt{2} 	imes h = \sqrt{2}$,so $h = 1$.The height $h$ is the perpendicular distance from point $C$ to the line segment $AB$. Since the radius of the circle is $r = \frac{AB}{2} = \sqrt{2} \approx 1.414$,and $h = 1 Each of these two lines intersects the circle at two distinct points. Therefore,there are $4$ possible positions for point $C$.

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 the area of $\Delta ABC = \sqrt{2}$ sq. unit,then the maximum number of positions of $C$ in the $xy$ plane is

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