Question

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

• A. $2\sqrt{3}$ sq. units
• B. $3\sqrt{2}$ sq. units
• C. $\sqrt{6}$ sq. units
• D. none of these

Answer 1

The correct option is A $2\sqrt{3}$ sq. units

$x^2+y^2=4,C=(0,0),r=2$

Let $y=mx+r\sqrt{m^2+1}$ be a tangent

$y=mx+2\sqrt{m^2+1}$

This passes through $(1,\sqrt{3})$

$\sqrt{3}=m+2\sqrt{m^2+1}$

$(\sqrt{3}-m{)}^2=4(m^2+1)$

$m^2+3–2\sqrt{3}m=4m^2+4$

$3m^2+2\sqrt{3}m+1=0$

$(m+\frac{1}{\sqrt{3}}{)}^2=0⟹m=\frac{-1}{\sqrt{3}}$

Equation is

$y=\frac{-x}{\sqrt{3}}+dfrac4\sqrt{3}$

$x+\sqrt{3}y=4$

Equation of normal will be

$\sqrt{3}x–y+c=0$

But normal passes through $(1,\sqrt{3})$

$C=0$

$\sqrt{3}x–y$ is equation of normal

Tangent, normal and x-axis form a triangle with vertices at (0,0) ,(4,0),$(1,\sqrt{3})$

$Area=\frac{1}{2}\times 4\times \sqrt{3}=2\sqrt{3}$