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:

  • A
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$

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