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. $150\text{sq. units}$
• B. $50\text{sq. units}$
• C. $75\text{sq. units}$
• D. $70\text{sq. units}$

Answer 1

The correct option is C $75\text{sq. units}$

Equation of tangent at $B(1,7)$ is,
$x+7y-(x+1)-2(y+7)-20=0\Rightarrow y=7$
Equation of tangent at $D(4,-2)$ is,
$4x-2y-(x+4)-2(y-2)-20=0\Rightarrow 3x-4y=20$

The coordinate of $C$ will be,
$=(16,7)$
So,
$BC=15$

Area of quadilateral $ABCD$ is,
$=2\times \frac{1}{2}\times 5\times 15=75\text{sq. units}$