Question

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

• A. $150$sq. units
• B. $75$sq. units
• C. $25$sq. units
• D. $100$sq. units

Answer 1

The correct option is B $75$sq. units

Equation of tangent at $B(1,7)$ and $D(4,-2)$ are

$x+7y-(x+1)-2(y+7)-20=0\Rightarrow 5y=35\Rightarrow y=7$

and $4x-2y-(x+4)-2(y-2)-20=0\Rightarrow 3x-4y=20$

Therefore point C is $(16,7)$

Therefore the area of quadrilateral $ABCD$

$=$area of $\Delta ABC+$area of $\Delta ACD$

$=\frac{1}{2}\times 15\times 5+\frac{1}{2}\times 15\times 5=75$