Question:
The tangent and the normal lines at the point \(\left(\sqrt{3}, 1\right)\) to the circle \(x^2 + y^2 = 4\) and the x-axis form a triangle. The area of this triangle (in square units) is :
The tangent and the normal lines at the point \(\left(\sqrt{3}, 1\right)\) to the circle \(x^2 + y^2 = 4\) and the x-axis form a triangle. The area of this triangle (in square units) is :
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The normal passes through the centre (origin); find where the tangent cuts the x-axis and use the triangle formed with the origin and the point.
Updated On: Nov 5, 2026
- \(\frac{4}{\sqrt{3}}\)
- \(\frac{1}{\sqrt{3}}\)
- \(\frac{2}{\sqrt{3}}\)
- \(\frac{1}{3}\)
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The Correct Option is C
Solution and Explanation
Tangent at \((\sqrt3, 1)\): \(\sqrt3 x + y = 4\). It meets the x-axis at \(A\left(\frac{4}{\sqrt3}, 0\right)\).
The normal at that point passes through the centre \(O(0,0)\): \(y = \frac{x}{\sqrt3}\).
The triangle has vertices \(O(0,0)\), \(A\left(\frac{4}{\sqrt3}, 0\right)\), \(P(\sqrt3, 1)\). Base \(OA = \frac{4}{\sqrt3}\), height = y-coordinate of \(P\) = 1.
Area \(= \frac12 \times \frac{4}{\sqrt3} \times 1 = \frac{2}{\sqrt3}\).
Answer: (3) \(\frac{2}{\sqrt{3}}\)
The normal at that point passes through the centre \(O(0,0)\): \(y = \frac{x}{\sqrt3}\).
The triangle has vertices \(O(0,0)\), \(A\left(\frac{4}{\sqrt3}, 0\right)\), \(P(\sqrt3, 1)\). Base \(OA = \frac{4}{\sqrt3}\), height = y-coordinate of \(P\) = 1.
Area \(= \frac12 \times \frac{4}{\sqrt3} \times 1 = \frac{2}{\sqrt3}\).
Answer: (3) \(\frac{2}{\sqrt{3}}\)
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