The equation of the normal to the circle x^2+ | vedclass

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(C) The equation of the circle is $x^2+y^2=16$,which has its center at $C(0,0)$.Any normal to a circle always passes through its center.Therefore,the

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$x-y=0$

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(C) The equation of the circle is $x^2+y^2=16$,which has its center at $C(0,0)$.Any normal to a circle always passes through its center.Therefore,the normal at the point $P\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ is the line passing through $C(0,0)$ and $P\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.The slope $m$ of the line $CP$ is given by $m = \frac{\frac{1}{\sqrt{3}} - 0}{\frac{1}{\sqrt{3}} - 0} = 1$.The equation of the line passing through $(0,0)$ with slope $m=1$ is $y - 0 = 1(x - 0)$,which simplifies to $x - y = 0$.Thus,option $C$ is correct.

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

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