Two tangents to the circle x^2+y^2=4 at the p | vedclass

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(C) The equation of the circle is $x^2+y^2=2^2$,so the radius $r=2$ and the center $O=(0,0)$.Point $P$ is $(-4,0)$. The distance $OP = 4$.In the right

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$4 \sqrt{3}$ sq. units

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(C) The equation of the circle is $x^2+y^2=2^2$,so the radius $r=2$ and the center $O=(0,0)$.Point $P$ is $(-4,0)$. The distance $OP = 4$.In the right-angled triangle $	riangle OAP$ (where $\angle OAP = 90^\circ$ because $PA$ is a tangent),$OA = r = 2$ and $OP = 4$.Using the Pythagorean theorem,$AP = \sqrt{OP^2 - OA^2} = \sqrt{4^2 - 2^2} = \sqrt{16-4} = \sqrt{12} = 2\sqrt{3}$.The area of $	riangle OAP = \frac{1}{2} 	imes OA 	imes AP = \frac{1}{2} 	imes 2 	imes 2\sqrt{3} = 2\sqrt{3}$ sq. units.Since the quadrilateral $PAOB$ is composed of two congruent triangles $	riangle OAP$ and $	riangle OBP$,the total area is $2 	imes 	ext{Area}(	riangle OAP) = 2 	imes 2\sqrt{3} = 4\sqrt{3}$ sq. units.

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

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