TS EAMCET · Maths · Circle
If \(\Delta\) is the area of the triangle formed by the positive \(x\)-axis and the normal and tangent to the circle \(x^2+y^2=4\) at \((1, \sqrt{3})\), then \(\Delta\) is equal to
- A \(\frac{\sqrt{3}}{2}\)
- B \(\sqrt{3}\)
- C \(2 \sqrt{3}\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
Given equation of circle is \(x^2+y^2=4\) On differentiating w.r.t. \(x\), we get…
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