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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Aicharya · Accepted Answer

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 $\triangle O A B$ $\begin{aligned} & =\frac{1}{2} \times O B \times A D=\frac{1}{2} \times 4 \times \sqrt{3} \\ & =2 \sqrt{3}\end{aligned}$

If $\Delta$ is the area of the triangle formed by the positive $x$-axis and the normal and tangent to the…

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