Question:
<p>Two tangents to the circle x<sup>2</sup> + y<sup>2</sup> = 4 at the points A and B meet at point P(-4, 0). The area of the quadrilateral PAOB in sq. units, where O is origin, is</p>
<p style="display:inline"><span class="math-tex">\(4 \sqrt{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(6 \sqrt{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(4 \sqrt{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(8 \sqrt{3}\)</span></p>
Step-by-Step Solution
Key Concept: For a point P outside circle x² + y² = 4, tangents from P to the circle touch at points A and B. The quadrilateral PAOB has area = (1/2) × PA × PB × sin(∠APB), where PA = PB = √(OP² - r²) and ∠AOB can be found using cos(∠AOB) = r/OP. The quadrilateral area equals sum of two right triangles OAP and OBP.
<html><body><p><img alt="" data-imgur-src="dVUOTCl.png" height="99" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/1623761639-8hber9.jpg" width="187"/><br/>
Let S <span class="math-tex">$\equiv$</span> x<sup>2</sup> + y<sup>2</sup> - 4 = 0<br/>
Required area <span class="math-tex">$=r \sqrt{S_{1}}=2 \sqrt{12}=4 \sqrt{3}$</span></p></body></html>
Correct Answer: C