Question

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, O being the origin, is

• A. $2\sqrt{3}$ sq. units
• B. $8\sqrt{3}$ sq. units
• C. $4\sqrt{3}$ sq. units
• D. $6\sqrt{3}$ sq. units

Hint:

Logic Tip: The length of a tangent $L$ from a point $(x_1, y_1)$ to a circle $x^2 + y^2 = r^2$ is simply $L = \sqrt{x_1^2 + y_1^2 - r^2}$. Here, $L = \sqrt{(-4)^2 + 0 - 4} = \sqrt{12} = 2\sqrt{3}$. The area is immediately $r \times L = 2 \times 2\sqrt{3} = 4\sqrt{3}$.

Answer 1

The Correct Option is C

Concept:
The radius to the point of tangency is always perpendicular to the tangent line. This forms two congruent right-angled triangles, $\Delta OAP$ and $\Delta OBP$, where $O$ is the center of the circle. The total area of quadrilateral $PAOB$ is twice the area of one of these right triangles.
Step 1: Identify the properties of the given circle and points.
The equation of the circle is $x^2 + y^2 = 4$. Its center $O$ is $(0, 0)$ and its radius $r$ is $\sqrt{4} = 2$. Thus, $OA = OB = 2$. The external point $P$ is $(-4, 0)$.
Step 2: Find the lengths of the sides of right-angled triangle OAP.
In right $\Delta OAP$, the hypotenuse is $OP$. Calculate the distance $OP$: $$OP = \sqrt{(-4 - 0)^2 + (0 - 0)^2} = \sqrt{16} = 4$$ Using Pythagoras theorem to find the length of the tangent $AP$: $$OP^2 = OA^2 + AP^2$$ $$4^2 = 2^2 + AP^2$$ $$16 = 4 + AP^2 \implies AP^2 = 12 \implies AP = \sqrt{12} = 2\sqrt{3}$$
Step 3: Calculate the area of the quadrilateral.
The area of right $\Delta OAP$ is: $$\text{Area}(\Delta OAP) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OA \times AP$$ $$\text{Area}(\Delta OAP) = \frac{1}{2} \times 2 \times 2\sqrt{3} = 2\sqrt{3} \text{ sq. units}$$ Because $\Delta OAP$ and $\Delta OBP$ are congruent, the total area of the quadrilateral $PAOB$ is: $$\text{Area}(PAOB) = 2 \times \text{Area}(\Delta OAP)$$ $$\text{Area}(PAOB) = 2 \times 2\sqrt{3} = 4\sqrt{3} \text{ sq. units}$$