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
$x^2+y^2-2x-4y-20=0$
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 $ABCD$
$=2\times ar(△ABC)$
$=2\times \frac{1}{2}\times 15\times 5=75sq$ units