Step 1: Understanding the Question:
We need to find the area of the quadrilateral formed by the center of the circle, the external point from which tangents are drawn, and the points of contact of those tangents.
Step 2: Key Formula or Approach:
For a circle $x^2 + y^2 = r^2$ and an external point $P(x_1, y_1)$:
1. Area of quadrilateral $PAOB = 2 \times \text{Area of } \triangle PAO$.
2. In right-angled triangle $PAO$ (where $A$ is point of contact), $OA = r$ and $OP = \sqrt{x_1^2 + y_1^2}$.
3. Length of tangent $PA = \sqrt{OP^2 - OA^2} = \sqrt{x_1^2 + y_1^2 - r^2}$.
Step 3: Detailed Explanation:
Given circle is $x^2 + y^2 = 4$, so center $O = (0, 0)$ and radius $r = 2$.
The external point is $P = (-4, 0)$.
The distance $OP = \sqrt{(-4)^2 + 0^2} = 4$.
Since tangent is perpendicular to radius at point of contact, $\triangle PAO$ is right-angled at $A$.
By Pythagoras theorem in $\triangle PAO$:
\[ PA = \sqrt{OP^2 - OA^2} = \sqrt{4^2 - 2^2} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3} \]
Area of $\triangle PAO = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times PA \times OA$
\[ \text{Area}(\triangle PAO) = \frac{1}{2} \times 2\sqrt{3} \times 2 = 2\sqrt{3} \]
The area of quadrilateral $PAOB$ is twice the area of $\triangle PAO$:
\[ \text{Area}(PAOB) = 2 \times 2\sqrt{3} = 4\sqrt{3} \text{ sq. units} \]
Step 4: Final Answer:
The area of quadrilateral PAOB is $4\sqrt{3}$ sq. units.