Question

Area of the triangle formed by the tangent and normal at the point $(1,\sqrt{3})$ of circle $x^2+y^2=4$ and with positive $x$ axis is.

• A. $2\sqrt{3}$
• B. $4\sqrt{3}$
• C. $\sqrt{3}$
• D. None of these

Answer 1

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

The given equation of circle,

$x^2+y^2=4$

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

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

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

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

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

Now, equation of normal to the circle is

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

Slope of tangent from equation ( 1 ) is $\frac{-1}{\sqrt{3}}$

We know,

Slope of normal $\times$ Slope of tangent $=-1$

$\Rightarrow$ Slope of normal $=\frac{-1}{\left(\frac{-1}{\sqrt{3}}\right)}=\sqrt{3}$

Now, equation of normal is

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

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

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

From diagram,

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

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

$=\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{3}}(-8+16-\frac{(-1)}{2}-4)$

$=\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{3}}\times \frac{9}{2}$

$=2\sqrt{3}$