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Find the area of the triangle enclosed by the tangent and the normal drawn at the point (1, √3) on the circle x² + y² = 4, together with the x-axis.
- 3 sq. units
- 2√3 sq. units
- 3√2 sq. units
- 4 sq. units
Correct answer: 2√3 sq. units
Solution
At (1,sqrt3) on x^2+y^2=4 the tangent has slope -1/sqrt3 and meets the x-axis at (4,0); the normal passes through the centre (0,0). The triangle has vertices (0,0),(4,0),(1,sqrt3), so area = (1/2)*4*sqrt3 = 2*sqrt3 sq. units.
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