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

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(C) The circle is $x^2+y^2=2^2$,so the radius $r = 2$.The distance $OP = \sqrt{(-4-0)^2 + (0-0)^2} = 4$.In the right-angled triangle $	riangle PBO$,t

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$4 \sqrt{3}$ sq. units

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(C) The circle is $x^2+y^2=2^2$,so the radius $r = 2$.The distance $OP = \sqrt{(-4-0)^2 + (0-0)^2} = 4$.In the right-angled triangle $	riangle PBO$,the length of the tangent $PB = \sqrt{OP^2 - OB^2} = \sqrt{4^2 - 2^2} = \sqrt{16-4} = \sqrt{12} = 2\sqrt{3}$.The area of the quadrilateral $PAOB$ is the sum of the areas of $	riangle PBO$ and $	riangle PAO$.Since $	riangle PBO \cong 	riangle PAO$,the area of $PAOB = 2 	imes 	ext{Area}(	riangle PBO)$.Area of $	riangle PBO = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes PB 	imes OB = \frac{1}{2} 	imes 2\sqrt{3} 	imes 2 = 2\sqrt{3}$.Therefore,the area of the quadrilateral $PAOB = 2 	imes 2\sqrt{3} = 4\sqrt{3}$ sq. units.

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

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