Q92 / 160 · AP EAMCET 2020 (18 Sep Shift 2) · Mathematics

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
\(x-y=\frac{\sqrt{3}}{4}\)
Correct answer
C
\(x-y=0\)
D
\(x+y=\frac{\sqrt{3}}{4}\)

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.

Hence, the correct answer is option: 'C'.

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