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Let $A$ be the centre of the circle $x^{2}+y^{2}-2 x-4 y$ $20=0$, and $B(1,7)$ and $D(4,-2)$ are points on the circle then, if tangents be drawn at $\mathrm{B}$ and $D$, which meet at $C$, then area of quadrilateral $A B C D$ is -
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The correct answer is:
75
Here, centre is $A(1,2)$, and Tangent at $B$ $(1,7)$ is
$x .1+y .7-1(x+1)-2(y+7)-20=0$
or $y=7$
$. .(1)$
Tangent at $D(4,-2)$ is
$$
3 x-4 y-20=0
$$
Solving (1) and (2), we get $C$ is (16,7)
Area $A B C D=2$ (Area of $\triangle \mathrm{ABC}$ )
$$
\begin{array}{l}
=2 \times \frac{1}{2} A B \times B C \\
A B \times B C=5 \times 15=75 \text { units }
\end{array}
$$
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