Two tangents to the circle x^{2}+y^{2}=4 at t | vedclass

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(A) The circle is $x^{2}+y^{2}=4$, so its radius $OA = 2$ and the center $O$ is $(0,0)$.Since $MA$ is a tangent, $\angle OAM = 90^{\circ}$.In the righ

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$4 \sqrt{3}$ sq. units

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(A) The circle is $x^{2}+y^{2}=4$, so its radius $OA = 2$ and the center $O$ is $(0,0)$.Since $MA$ is a tangent, $\angle OAM = 90^{\circ}$.In the right-angled triangle $	riangle OAM$, the hypotenuse $OM = 4$ and the side $OA = 2$.Using the Pythagorean theorem, $MA = \sqrt{OM^{2} - OA^{2}} = \sqrt{4^{2} - 2^{2}} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3}$.The quadrilateral $MAOB$ consists of two congruent right-angled triangles, $	riangle OAM$ and $	riangle OBM$.Therefore, the area of quadrilateral $MAOB = 2 	imes 	ext{Area}(	riangle OAM)$.Area $(	riangle OAM) = \frac{1}{2} 	imes OA 	imes MA = \frac{1}{2} 	imes 2 	imes 2\sqrt{3} = 2\sqrt{3}$.Thus, the area of quadrilateral $MAOB = 2 	imes 2\sqrt{3} = 4\sqrt{3}$ sq. units.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

Two tangents to the circle $x^{2}+y^{2}=4$ at the points $A$ and $B$ meet at $M(-4,0)$. The area of the quadrilateral $MAOB$, where $O$ is the origin, is

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