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. $2\sqrt{3}$ 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

The given equation of circle is $x^2+y^2=4$

The equation of the normal to the circle at $(1,\sqrt{3})$ is same as the line joining the points $(1,\sqrt{3})$ and $(0,0)$, which is given by

$\frac{y-\sqrt{3}}{x}-1=\frac{\sqrt{3}-0}{1-0}$

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

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

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

So, the slope of normal is $\sqrt{3}.$

We know that the product of the slope of the tangent and normal is $-1.$

$\therefore$ The slope of tangent is $\frac{-1}{\sqrt{3}}$

Now, the equation of the tangent to the circle at $(1,\sqrt{3})$ is given by

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

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

$\Rightarrow$ $y=\frac{-(x+4)}{\sqrt{3}}$ ----- ( 2 )

Putting $y=0$ in ( 2 ), we get $x=4$.

Thus, $ABC$ is triangle formed by the positive $x-axis$ and tangent and normal to the given circle at $(1,\sqrt{3})$

Now,

Area of $△AOB=$ Area of $△AOM+$ Area of $△AMB$

$={\int }_0^{1}ydx+{\int }_1^{4}ydx$

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

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

$=(\frac{\sqrt{3}}{2}-0)-{\left[\frac{x^2}{2\sqrt{3}}\right]}_1^4+{\left[\frac{4}{\sqrt{3}}x\right]}_1^4$

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

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

$=2\sqrt{3}$