If the circles ${x^2} + {y^2} + 2x + 2ky + 6 = 0$ and ${x^2} + {y^2} + 2ky + k = 0$ intersect orthogonally, then $k$ is

  • [IIT 2000]
  • A

    $2$ or $ - \frac{3}{2}$

  • B

    $-2$ or $\frac{3}{2}$

  • C

    $2$ or $\frac{3}{2}$

  • D

    -$2$ or -$\frac{3}{2}$

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