If Delta is the area of the triangle formed b | vedclass

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(C) Given the equation of the circle is $x^2+y^2=4$.On differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}

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$2 \sqrt{3}$

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(C) Given the equation of the circle is $x^2+y^2=4$.On differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{y}$.At the point $(1, \sqrt{3})$,the slope of the tangent is $m_t = -\frac{1}{\sqrt{3}}$.The equation of the tangent at $(1, \sqrt{3})$ is $y - \sqrt{3} = -\frac{1}{\sqrt{3}}(x - 1)$,which simplifies to $x + \sqrt{3}y = 4$.The tangent intersects the positive $x$-axis at point $B(4, 0)$.The slope of the normal at $(1, \sqrt{3})$ is $m_n = \sqrt{3}$.The equation of the normal at $(1, \sqrt{3})$ is $y - \sqrt{3} = \sqrt{3}(x - 1)$,which simplifies to $\sqrt{3}x - y = 0$.The normal passes through the origin $O(0, 0)$.The triangle is formed by the points $O(0, 0)$,$B(4, 0)$,and $A(1, \sqrt{3})$.The area of $	riangle OAB$ is given by $\Delta = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes OB 	imes AD$,where $AD$ is the $y$-coordinate of point $A$.$\Delta = \frac{1}{2} 	imes 4 	imes \sqrt{3} = 2\sqrt{3}$.

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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