Question:
<p>The tangents at B(1, 7) and D(4, -2) to the circle <span class="math-tex">\(\omega\)</span> : x<sup>2</sup> + y<sup>2</sup> - 2x - 4y - 20 = 0 meet at C. If the centre of the circle is A, then the area of <span class="math-tex">\(\square\)</span>ABCD (in sq. units) is</p>
<p style="display:inline">125</p>
<p style="display:inline">75</p>
<p style="display:inline">150</p>
<p style="display:inline"><span class="math-tex">\(\frac{75}{2}\)</span></p>
Step-by-Step Solution
Key Concept: The area of quadrilateral ABCD is calculated as twice the area of the right-angled triangle formed by the radius and the tangent, effectively becoming the product of the radius and the tangent length.
<html><body><p><img alt="" data-imgur-src="yOXOdWu.png" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/1623761549-a9cga2.jpg" style="width: 171px; height: 113px;"/><br/>
<span class="math-tex">$\omega$</span> : (x - 1)<sup>2</sup> + (y - 2)<sup>2</sup> = (5)<sup>2</sup><br/>
<span class="math-tex">$\Rightarrow$</span> Centre A = (1, 2), radius = 5<br/>
Slope of AD <span class="math-tex">$=\frac{2-(-2)}{1-4}=-\frac{4}{3}$</span><br/>
Slope of AB is not defined and equation of AB is given by x = 1.<br/>
Equation of CD is<br/>
3x - 4y - 20 = 0 ...(i)<br/>
Equation of BC is y = 7 ...(ii)<br/>
Solving (i) and (ii), we get C = (16, 7)<br/>
Area of <span class="math-tex">$\square$</span>ABCD = 2 (area of <span class="math-tex">$\triangle$</span>ABC)<br/>
<span class="math-tex">$=2\left(\frac{1}{2} \times 5 \times 15\right)$</span> = 75</p></body></html>
Correct Answer: B