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Let $A$ be the centre of the circle $x^{2}+y^{2}-2 x-4 y-20=0 .$ 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 $A B C D$ is
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$D(4,-2)$ be two points on the circle such that tangents at $B$ and $D$ meet at $C$. The area of the quadrilateral $A B C D$ is
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Verified Answer
The correct answer is:
$75 \mathrm{sq}$ units
Given, equation of circle is
$$
x^{2}+y^{2}-2 x-4 y-20=0
$$
Center $\left(1,2\right).$ and radius $=\sqrt{(1)^{2}+\left(2)^{2}+20\right.}=5$
Coordinate of intersecting point of tangents at $B$ and $D$ is $C(06,7)$
Area of quadrilateral $A B C D$
$$
=2 \times \operatorname{ar}(\Delta A B C)
$$
$=2 \times \frac{1}{2} \times 15 \times 5=75 \mathrm{sq}$ units
$$
x^{2}+y^{2}-2 x-4 y-20=0
$$
Center $\left(1,2\right).$ and radius $=\sqrt{(1)^{2}+\left(2)^{2}+20\right.}=5$
Coordinate of intersecting point of tangents at $B$ and $D$ is $C(06,7)$
Area of quadrilateral $A B C D$
$$
=2 \times \operatorname{ar}(\Delta A B C)
$$
$=2 \times \frac{1}{2} \times 15 \times 5=75 \mathrm{sq}$ units
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