Question

Two tangents to $x^{2}+y^{2}=4$ at A and B meet at $P(-4,0).$ Area of quadrilateral PAOB is}

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

Hint:

Length of tangent $L = \sqrt{S_{1}}$. Area of quad $= rL$.

Answer 1

The Correct Option is C

Step 1: Concept
Tangents from an external point are perpendicular to the radii at the points of tangency.

Step 2: Analysis
$O=(0,0)$, radius $r=2$. $P=(-4,0)$, so $OP=4$.
In right $\triangle PAO$, $PA = \sqrt{OP^{2} - r^{2}} = \sqrt{4^{2} - 2^{2}} = \sqrt{12} = 2\sqrt{3}$.

Step 3: Calculation
Area of $\triangle PAO = \frac{1}{2} \times PA \times r = \frac{1}{2} \times 2\sqrt{3} \times 2 = 2\sqrt{3}$.
Area of Quad PAOB $= 2 \times Area(\triangle PAO) = 4\sqrt{3}$.

Step 4: Conclusion
Hence, the area is $4\sqrt{3}$ sq. units. Final Answer: (C)