The equation of the normal to the circle $x^2+y^2=16$ at the point $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ is

  • A
    $x+y=0$
  • B
    $x-y=\frac{\sqrt{3}}{4}$
  • C
    $x-y=0$
  • D
    $x+y=\frac{\sqrt{3}}{4}$

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