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

Question

Dr. Abubacker · Accepted Answer

(a)
In $\mathrm{\Delta}POB$,
[image:11_image21.emf]
$\sin	heta = \frac{2}{4} = \frac{1}{2}$
$\Rightarrow \ \ \ 	heta = 30{^\circ}$
$	herefore\ \ area(\mathrm{\Delta}POA) = \frac{1}{2} 	imes 2 	imes 4 	imes \sin{30{^\circ}} = 2$
Hence, area (quad $PAOB$)=2 area $(\mathrm{\Delta}POA)$
$= 2 	imes 2 = 4$ sq units

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

Explanation

Reviewed by Dr. Abubacker

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