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

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$2\sqrt{3}$

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(A) The equation of the circle is $x^2 + y^2 = 4$. The point of contact is $P(1, \sqrt{3})$.The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$. Substituting the values,we get $x(1) + y(\sqrt{3}) = 4$,which simplifies to $x + \sqrt{3}y = 4$.The tangent intersects the $x$-axis at $y = 0$,so $x = 4$. The point is $A(4, 0)$.The normal at $(1, \sqrt{3})$ passes through the center $(0, 0)$ and the point $(1, \sqrt{3})$. Its equation is $y = \sqrt{3}x$.The normal intersects the $x$-axis at the origin $O(0, 0)$.The triangle is formed by the vertices $O(0, 0)$,$P(1, \sqrt{3})$,and $A(4, 0)$.The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area $= \frac{1}{2} |0(\sqrt{3} - 0) + 1(0 - 0) + 4(0 - \sqrt{3})| = \frac{1}{2} |-4\sqrt{3}| = 2\sqrt{3}$ square units.

The area of the triangle formed by the tangent and the normal drawn at $(1, \sqrt{3})$ to the circle $x^2 + y^2 = 4$ and the positive $x$-axis is:

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