Question

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

• A. $\sqrt[2]{21}$ sq.units
• B. $\sqrt[2]{22}$ sq.units
• C. $\sqrt{6}$ sq.units
• D. $\sqrt[2]{212}$ sq.units

Answer 1

Answer: (D)

$\sqrt[2]{212}$ sq.units

Given, $x^2+y^2=4$
On differentiating w.r. to $x$, we get

$2x+2y.y^{\prime }=0$
$\therefore y^{\prime }=\frac{-x}{y}$

$\therefore {\frac{dy}{dx}}_{1,\sqrt{3}}=\frac{-1}{\sqrt{3}}$

Therefore equation of the tangent is

$y-\sqrt{3}=\frac{-1}{\sqrt{3}}(x-1)$

$\sqrt{3}y-3=1-x$

$x+\sqrt{3}y=4$ ...(i)

It cuts the $x$ axis at $x=4$.

Similarly equation of the normal will be

$(y-\sqrt{3})=\sqrt{3}(x-1)$

$y-\sqrt{3}=\sqrt{3}x-\sqrt{3}$

$\sqrt{3}x=y$ ...(ii)

Hence, the normal cuts the $x$ -axis at $(0,0)$.

Therefore the coordinates of the triangle are

$(0,0),(4,0),(1,\sqrt{3})$

$=\frac{4\sqrt{3}}{2}$ .... right angled triangle, right angled at $(1,\sqrt{3})$.

$=2\sqrt{3}$ sq units.