The equation of the normal to the circle \(x^2+y^2=16\) at the point \(\left(\frac{1}{\sqrt{3}},…
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
\(x+y=0\)
\(x-y=\frac{\sqrt{3}}{4}\)
\(x-y=0\)
\(x+y=\frac{\sqrt{3}}{4}\)
Solution
Centre \(C=(0,0)\)
\(P=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\)
Since, every normal passes through centre
\(\therefore\) Equation of Normal \(=\) Equation of \(\mathrm{CP}\)
\(\begin{array}{ll}
\Rightarrow & y-0=\frac{\frac{1}{\sqrt{3}}-0}{\frac{1}{\sqrt{3}}-0}(x-0) \Rightarrow y=1 \cdot x \\
\Rightarrow & x-y=0
\end{array}\)
Hence, option (c) is correct.