Question

Let $A$ be the centre of the circle $x^2+y^2-2x-4y-20=0$. If the tangents drawn at the point $B(1,7)$ and $D(4,-2)$ on the given circle meet at the point $C$, then the area of the quadrilateral $ABCD$ is

• A. 60
• B. 65
• C. 70
• D. 75

Answer 1

Answer: (D)

Equation of circle
$x^2+y^2-2x-4y-20=0$
Centre of circle $(1,2)$, radius $=5$

Equation of tangents at $(1,7)$ is
$(1)x+7y-(x+1)-2(y+7)-20=0$
$\Rightarrow x+7y-x-1-2y+7-20=0$
$\Rightarrow 5y=35$
$\Rightarrow y=7\dots (i)$
Equation of tangent at $(4,-2)$
$4x-2y-(x+4)-2(y-2)-20=0$
$\Rightarrow 4x-2y-x-4-2y+4-20=0$
$\Rightarrow 3x-2y=20\dots (ii)$
Tangents $y=7$ and $3x-2y=20$ intersect at $C$. So, coordinate of $C(16,7)$
Length of $BC=\sqrt{(16-1{)}^2+(7-7{)}^2}$
$BC=15$
Length of $CD=\sqrt{(16-4{)}^2+(7+2{)}^2}=\sqrt{144+81}$
$CD=15$
So, area of $ABCD=$ area of $\Delta ABC+$ area of $\Delta ACD$
$=\frac{1}{2}\times 15\times 5+\frac{1}{2}\times 15\times 5=75$