Question

The area of the region (in sq. unit) bounded by x-axis, the tangent and normal to the circle \(x^2+y^2 = 4\), drawn at a point \((1,\sqrt{3})\) is

• A. \(5\)
• B. \(\sqrt{3}\)
• C. \(2\)
• D. \(2\sqrt{3}\)

Hint:

Find the tangent and normal at the point, then the triangle they make with the x-axis.

Answer 1

The Correct Option is D

Step 1: Use the determinant formula:
Vertices $(0,0)$, $(4,0)$, $(1,\sqrt3)$.

Step 2: Compute:
Area $= \frac12\left|0(0 - \sqrt3) + 4(\sqrt3 - 0) + 1(0 - 0)\right| = \frac12 \times 4\sqrt3 = 2\sqrt3$.

Final Answer:
The area is $2\sqrt{3}$ sq. units, option (D). \[ \boxed{2\sqrt{3}} \]