Question

Let $A$ be the center of the circle $x^2+y^2-2x-4y-20=0$, and $B(1,7)$ and $D(4,-2)$ are points on the circle then, if tangents be drawn at $B$ and $D$, which meet at $C$, then area of quadrilateral $ABCD$ is

• A. 150
• B. 75
• C. 75/2
• D. None of these

Answer 1

Answer: (B)

$x^2+y^2-2x-4y-20=0$
$A=(1,2)$
Equation of tangent at $B(1,7)$
$x+7y-(x+1)-2(y+7)-20=0$
$5y=35$
$y=7$
Equation of tangent at $D(4,-2)$
$4x-2y-(x+4)-2(y-2)-20=0$
$3x-4y=20$
$\therefore$ coordinates of $C$ are $(16,7)$
Length of $AB=\sqrt{(1-1{)}^2+(7-2{)}^2}=5$
Length of $BC=\sqrt{(16-1{)}^2+(7-7{)}^2}=15$
$\therefore$ The area of quadrilateral
$ABCD=2\times \frac{1}{2}\times 5\times 15$
$=75$ sq. units