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 $(1,\sqrt{3})$ is

• A. $3\sqrt{2}sq$. unit
• B. $2\sqrt{3}sq$. unit
• C. $\sqrt{6}sq$. unit
• D. $Noneofthese$

Answer 1

Answer: (B)

$2\sqrt{3}sq$. unit

Consider the given equation.

$x^2+y^2=4$

We know that the equation of the tangent to the circle $x^2+y^2=C$ at point $(x_1,y_1)$ is is given by

$x{x}_1+y{y}_1=C$

So, the equation of the tangent at $(1,\sqrt{3})$ is

$x$ + $\sqrt{3}$ $y=4$ .......... (1)

$y=\frac{4}{\sqrt{3}}-\frac{x}{\sqrt{3}}$, it cuts the $x$ axis at $(4,0)$.

Now, equation of normal to the circle is

$(y-\sqrt{3})$ = Slope of normal $\times (x-1)$.

Slope of normal = $\frac{-1}{Slopeoftangent}$

So,

Slope of normal = $\sqrt{3}$

Now, equation of normal is

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

$y=\sqrt{3}x$ .......... (2)

Therefore, in figure 1, the area formed by tangent, normal and x axis will be,

$A={\int }_0^1\sqrt{3}xdx$ + ${\int }_1^4(\frac{-x}{\sqrt{3}}+\frac{4}{\sqrt{3}})dx$

$A={\sqrt{3}\left[\frac{x^2}{2}\right]}_0^1$ + ${\frac{1}{\sqrt{3}}[-\frac{x^2}{2}+4x]}_1^4$

$A=\frac{\sqrt{3}}{2}$ + $\frac{1}{\sqrt{3}}[-\frac{16}{2}+16+\frac{1}{2}-4]$

$A=\frac{\sqrt{3}}{2}$ + $\frac{1}{\sqrt{3}}\left[\frac{9}{2}\right]$

$A=\frac{\sqrt{3}}{2}+\frac{3\sqrt{3}}{2}$

$A=\frac{4\sqrt{3}}{2}$

$A=2\sqrt{3}$

Hence, option b is correct.