The area of the region (in sq. unit) bounded | vedclass

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

Vedclass · Accepted Answer

$2\sqrt{3}$

Vedclass · Answer

(D) $1$. The equation of the circle is $x^2 + y^2 = 4$. The point of contact is $P(1, \sqrt{3})$.$2$. The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$. Thus, $x(1) + y(\sqrt{3}) = 4$, or $x + \sqrt{3}y = 4$.$3$. The $x$-intercept of the tangent is found by setting $y=0$, giving $x=4$. So, the tangent meets the $x$-axis at $A(4, 0)$.$4$. The normal at any point on a circle passes through the center $(0, 0)$. The normal passes through $(0, 0)$ and $(1, \sqrt{3})$. Its equation is $y = \sqrt{3}x$.$5$. The normal meets the $x$-axis at the origin $O(0, 0)$.$6$. The region is a triangle with vertices $O(0, 0)$, $P(1, \sqrt{3})$, and $A(4, 0)$.$7$. The area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes |4 - 0| 	imes |\sqrt{3} - 0| = \frac{1}{2} 	imes 4 	imes \sqrt{3} = 2\sqrt{3}$ sq. units.

The area of the region (in sq. unit) bounded by the $x$-axis, the tangent, and the normal to the circle $x^2 + y^2 = 4$ drawn at the point $(1, \sqrt{3})$ is

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