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

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(C) The given 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 center $A$

Vedclass · Accepted Answer

$75 	ext{ sq units}$

Vedclass · Answer

(C) The given 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 center $A$ is $(-g, -f) = (1, 2)$ and the radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{1+4+20} = \sqrt{25} = 5$. The tangent at $B(1, 7)$ is $x(1) + y(7) - (x+1) - 2(y+7) - 20 = 0$, which simplifies to $y=7$. The tangent at $D(4, -2)$ is $x(4) + y(-2) - (x+4) - 2(y-2) - 20 = 0$, which simplifies to $3x-4y-20=0$. Solving $y=7$ and $3x-4y-20=0$, we get $3x-28-20=0 \implies 3x=48 \implies x=16$. Thus, $C$ is $(16, 7)$. The quadrilateral $ABCD$ consists of two congruent triangles $\Delta ABC$ and $\Delta ADC$. The length of the tangent $BC = \sqrt{(16-1)^{2} + (7-7)^{2}} = 15$. The area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes BC 	imes AB = \frac{1}{2} 	imes 15 	imes 5 = 37.5$. The area of quadrilateral $ABCD = 2 	imes 	ext{Area}(\Delta ABC) = 2 	imes 37.5 = 75 	ext{ sq units}$.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

Let $A$ be the centre of the circle $x^{2}+y^{2}-2x-4y-20=0$. 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 $ABCD$ is

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