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If $\Delta$ is 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 $(1, \sqrt{3})$, then $\Delta$ is equal to
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The correct answer is:
$2 \sqrt{3}$
Given equation of circle is
$x^2+y^2=4$
On differentiating w.r.t. $x$, we get
$\begin{aligned} & 2 x+2 y \frac{d y}{d x}=0 \\ & \Rightarrow \quad \frac{d y}{d x}=-\frac{x}{y} \\ & \Rightarrow \quad\left(\frac{d y}{d x}\right)_{(1, \sqrt{3})}=-\frac{1}{\sqrt{3}} \\ & \end{aligned}$
$\therefore$ Equation of tangent at $(1, \sqrt{3})$ is
$\begin{aligned} & & y-\sqrt{3} & =-\frac{1}{\sqrt{3}}(x-1) \\ \Rightarrow & & \sqrt{3} y-3 & =-x+1 \\ \Rightarrow & & x+\sqrt{3} y & =4\end{aligned}$
and equation of normal at $(1, \sqrt{3})$ is
$\begin{array}{rlrl}y-\sqrt{3} & =\sqrt{3}(x-1) \\ \Rightarrow \quad & & \sqrt{3} x-y & =0\end{array}$
$\therefore$ Intersection point of Eqs. (i) and (ii) is $(1, \sqrt{3})$.
$\therefore$ Area of $\triangle O A B$
$\begin{aligned} & =\frac{1}{2} \times O B \times A D=\frac{1}{2} \times 4 \times \sqrt{3} \\ & =2 \sqrt{3}\end{aligned}$
$x^2+y^2=4$
On differentiating w.r.t. $x$, we get
$\begin{aligned} & 2 x+2 y \frac{d y}{d x}=0 \\ & \Rightarrow \quad \frac{d y}{d x}=-\frac{x}{y} \\ & \Rightarrow \quad\left(\frac{d y}{d x}\right)_{(1, \sqrt{3})}=-\frac{1}{\sqrt{3}} \\ & \end{aligned}$
$\therefore$ Equation of tangent at $(1, \sqrt{3})$ is
$\begin{aligned} & & y-\sqrt{3} & =-\frac{1}{\sqrt{3}}(x-1) \\ \Rightarrow & & \sqrt{3} y-3 & =-x+1 \\ \Rightarrow & & x+\sqrt{3} y & =4\end{aligned}$
and equation of normal at $(1, \sqrt{3})$ is
$\begin{array}{rlrl}y-\sqrt{3} & =\sqrt{3}(x-1) \\ \Rightarrow \quad & & \sqrt{3} x-y & =0\end{array}$
$\therefore$ Intersection point of Eqs. (i) and (ii) is $(1, \sqrt{3})$.
$\therefore$ Area of $\triangle O A B$
$\begin{aligned} & =\frac{1}{2} \times O B \times A D=\frac{1}{2} \times 4 \times \sqrt{3} \\ & =2 \sqrt{3}\end{aligned}$
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