AP EAMCET · Maths · Circle
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 \(x+y=\frac{\sqrt{3}}{4}\)
Answer & Solution
Correct Answer
(C) \(x-y=0\)
Step-by-step Solution
Detailed explanation
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 \(\mathrm{CP}\)…
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