The area (in sq units) of the triangle formed by the tangent, normal at $(1, \sqrt{3})$ to the circle $x^2+y^2=4$ and the $X$-axis, is

• A

  $4 \sqrt{3}$
• B

  $\frac{7}{2} \sqrt{3}$
• Correct answer
  C

  $2 \sqrt{3}$
• D

  $\frac{1}{2} \sqrt{3}$

Solution

The equations of the tangent and normal to the circle $x^2+y^2=4$ at $P(1, \sqrt{3})$ are $$ x+\sqrt{3} y=4 \text { and } y=\sqrt{3} x $$ The tangent meets $X$-axis at $A(4,0)$.
$\therefore$ Area of $\triangle O A P=\frac{1}{2} \times 4 \times \sqrt{3}=2 \sqrt{3}$ sq units

Hence, the correct answer is option: 'C'.