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(B) The equation of the circle is $x^2+y^2=4$. The point $P(\sqrt{3}, 1)$ lies on the circle.The equation of the tangent at $P(x_1, y_1)$ is $xx_1+yy_

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

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(B) The equation of the circle is $x^2+y^2=4$. The point $P(\sqrt{3}, 1)$ lies on the circle.The equation of the tangent at $P(x_1, y_1)$ is $xx_1+yy_1=r^2$,which gives $\sqrt{3}x+y=4$.For the tangent,the $X$-intercept is found by setting $y=0$,giving $x = \frac{4}{\sqrt{3}}$. So the point is $A(\frac{4}{\sqrt{3}}, 0)$.The normal at any point on a circle passes through the center $(0, 0)$. The line passing through $(0, 0)$ and $(\sqrt{3}, 1)$ is $y = \frac{1}{\sqrt{3}}x$,or $x - \sqrt{3}y = 0$.The $X$-intercept of the normal is at the origin $O(0, 0)$.The triangle is formed by the vertices $O(0, 0)$,$A(\frac{4}{\sqrt{3}}, 0)$,and $P(\sqrt{3}, 1)$.The base of the triangle along the $X$-axis is $OA = \frac{4}{\sqrt{3}}$.The height of the triangle is the $y$-coordinate of $P$,which is $1$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{4}{\sqrt{3}} 	imes 1 = \frac{2}{\sqrt{3}}$ sq. units.

If the tangent and the normal at the point $(\sqrt{3}, 1)$ to the circle $x^2+y^2=4$,and the $X$-axis form a triangle,then the area (in sq. units) of this triangle is

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