In $\triangle ABC$,$\angle A = 90^{\circ}$ and the coordinates of points $B$ and $C$ are $(2, -4)$ and $(1, 5)$. Then the equation of the circumcircle of $\triangle ABC$ is

  • A
    $x^2+y^2+3x+y+18=0$
  • B
    $x^2+y^2-3x+y-18=0$
  • C
    $x^2+y^2-3x-y-18=0$
  • D
    $x^2+y^2+3x-y+18=0$

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