Question

Let $A$ be the centre of the circle $x^2 + y^2 - 2x - 4y - 20 = 0$. Let $B (1, 7)$ and $D (4, -2)$ be two points on the circle such that tangents at $B$ and $D$ meet at $C$. The area of the quadrilateral $ABCD$ is

• A. 150 sq. units
• B. 50 sq. units
• C. 75 sq. units
• D. 70 sq. units

Answer 1

Answer: (C)

Given, equation of circle is
$x^{2}+y^{2}-2 x-4 y-20=0 $
Center $(1,2)$ and radius $=\sqrt{(1)^{2}+(2)^{2}+20}=5$

Coordinate of intersecting point of tangents at $B$ and $D$ is $C(16,7)$
$\therefore $ Area of quadrilateral $A B C D$
$=2 \times ar (\triangle A B C) $
$=2 \times \frac{1}{2} \times 15 \times 5=75\, sq$ units