Let A be the centre of the circle x^2+y^2-2x- | vedclass

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(D) The equation of the circle is $x^2+y^2-2x-4y-20=0$.Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=-2, c=-20$.The centre $A$ is $(-g, -f) = (

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$75$

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(D) The equation of the circle is $x^2+y^2-2x-4y-20=0$.Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=-2, c=-20$.The centre $A$ is $(-g, -f) = (1, 2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{1+4+20} = \sqrt{25} = 5$.The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 - (x+x_1) - 2(y+y_1) - 20 = 0$.For point $B(1, 7)$,the tangent is $x(1) + y(7) - (x+1) - 2(y+7) - 20 = 0$,which simplifies to $x + 7y - x - 1 - 2y - 14 - 20 = 0$,so $5y = 35$,or $y = 7$.For point $D(4, -2)$,the tangent is $x(4) + y(-2) - (x+4) - 2(y-2) - 20 = 0$,which simplifies to $4x - 2y - x - 4 - 2y + 4 - 20 = 0$,so $3x - 4y = 20$.Substituting $y = 7$ into $3x - 4y = 20$,we get $3x - 4(7) = 20$,so $3x = 48$,which gives $x = 16$.Thus,the point of intersection $C$ is $(16, 7)$.The quadrilateral $ABCD$ consists of two congruent right-angled triangles $	riangle ABC$ and $	riangle ADC$,where the right angle is at $B$ and $D$ respectively.The area of $	riangle ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes BC$.$AB = r = 5$. $BC = \sqrt{(16-1)^2 + (7-7)^2} = 15$.Area of $	riangle ABC = \frac{1}{2} 	imes 5 	imes 15 = 37.5$.Since $	riangle ADC \cong 	riangle ABC$,the total area of quadrilateral $ABCD = 2 	imes 37.5 = 75$.

Let $A$ be 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 given circle meet at the point $C$,then the area of the quadrilateral $ABCD$ is

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