Question

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

• A. $45sq.units$
• B. $95sq.units$
• C. $65sq.units$
• D. $75sq.units$

Answer 1

The correct option is D $75sq.units$

Circle is $(1,2),5$
$T_A$ is $3x-4y-20=0$, $T_B$ is $y-7=0$
Both meet at $C(16,7)\therefore CB=CD=15$ by distance formula
or $CB=CD=t=\sqrt{S^{\prime }}=\sqrt{225}=15$
Area of quadrilateral
$=2△ABC=2\cdot \frac{1}{2}\cdot r\cdot t=5\times 15=75sq.units.$