Let A(2, 3),B(4, 5) and let C = (x, y) be a p | vedclass

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

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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 $A(2, 3)$ and $B(4, 5)$ as the endpoints of a diameter.The length of the diameter $AB$ is $\sqrt{(4 - 2)^2 + (5 - 3)^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$.The area of $\Delta ABC$ is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \sqrt{2}$.Taking $AB$ as the base,the length of the base is $2\sqrt{2}$.So,$\frac{1}{2} 	imes (2\sqrt{2}) 	imes h = \sqrt{2}$,which simplifies to $h = 1$.The height $h$ represents the perpendicular distance from point $C$ to the diameter $AB$.Since the radius of the circle is $r = \frac{AB}{2} = \sqrt{2} \approx 1.414$,and the required height $h = 1$ is less than the radius,there are two chords parallel to $AB$ at a distance of $1$ unit on either side of $AB$.Each of these two chords intersects the circle at two distinct points.Therefore,there are $2 + 2 = 4$ possible positions for point $C$.

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

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