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

The circle $x^2 + y^2 = 4$ has center $O(0,0)$ and radius $2$ units.
The external point is $P(-4,0)$ with $OP = 4$ units.

Tangents from $P$ touch the circle at $A$ and $B$, forming right angles with radii $OA$ and $OB$.
The quadrilateral $PAOB$ consists of two congruent right triangles.

Using the Pythagorean theorem in $\triangle OAP$:
$PA^2 = OP^2 - OA^2 = 16 - 4 = 12$, so $PA = 2\sqrt{3}$ units.

The area of $\triangle OAP$ is $\frac{1}{2} \cdot OA \cdot PA = \frac{1}{2} \cdot 2 \cdot 2\sqrt{3} = 2\sqrt{3}$ sq. units.

Area of quadrilateral $PAOB = 2 \times 2\sqrt{3} = 4\sqrt{3}$ sq. units.