JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
The tangent and the normal lines at the point \((\sqrt 3,1)\) to the circle \(x^2 + y^2 = 4\) and the \(x -\) axis form a triangle. The area of this triangle (in square units) is
- A \(\frac{1}{{\sqrt 3 }}\)
- B \(\frac{4}{{\sqrt 3 }}\)
- C \(\frac{1}{3}\)
- D \(\frac{2}{{\sqrt 3 }}\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{{\sqrt 3 }}\)
Step-by-step Solution
Detailed explanation
Slope of \(OP = \frac{1}{{\sqrt 3 }}\) Equation of \(PQ\) is \(y - 1 = - \sqrt 3 \left( {x - \sqrt 3 } \right)\) \( \Rightarrow y + \sqrt 3 x = 4\) \( \Rightarrow Q\left( {\frac{4}{{\sqrt 3 }},0} \right)\) Area \( = \frac{2}{{\sqrt 3 }}\)
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