A is the centre of the circle x^2+y^2-2x-4y-2 | vedclass

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(A) The equation of the given circle is $x^2+y^2-2x-4y-20=0$ ... $(i)$.The centre $A$ is $(1, 2)$ and the radius $r = \sqrt{1^2+2^2+20} = \sqrt{25} =

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$75$

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(A) The equation of the given circle is $x^2+y^2-2x-4y-20=0$ ... $(i)$.The centre $A$ is $(1, 2)$ and the radius $r = \sqrt{1^2+2^2+20} = \sqrt{25} = 5$.The equation of the tangent at $B(1, 7)$ is $x(1) + y(7) - (x+1) - 2(y+7) - 20 = 0$,which simplifies to $5y = 35$,or $y = 7$ ... $(ii)$.The equation of the tangent at $D(4, -2)$ is $x(4) + y(-2) - (x+4) - 2(y-2) - 20 = 0$,which simplifies to $3x - 4y = 20$ ... $(iii)$.Solving $(ii)$ and $(iii)$ for the intersection point $C$: Substituting $y=7$ into $(iii)$,$3x - 4(7) = 20$ $\Rightarrow 3x = 48$ $\Rightarrow x = 16$. So,$C = (16, 7)$.The area of the quadrilateral $ABCD$ is $2 	imes 	ext{Area}(	riangle ABC) = 2 	imes (\frac{1}{2} 	imes r 	imes L)$,where $L$ is the length of the tangent from $C$ to the circle.$L = \sqrt{S_1} = \sqrt{16^2 + 7^2 - 2(16) - 4(7) - 20} = \sqrt{256 + 49 - 32 - 28 - 20} = \sqrt{225} = 15$.Area $= r 	imes L = 5 	imes 15 = 75$ square units.Thus,option $(A)$ is correct.

$A$ is the centre of the circle $x^2+y^2-2x-4y-20=0$. If the tangents drawn at the points $B(1,7)$ and $D(4,-2)$ on the circle meet at the point $C$,then the area of the quadrilateral $ABCD$ (in square units) is

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