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

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

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) The equation of the circle is $x^2 + y^2 - 2x - 4y - 20 = 0$. The center $A$ is $(1, 2)$ and the radius $r$ is $\sqrt{1^2 + 2^2 - (-20)} = \sqrt{1 + 4 + 20} = 5$.The tangent at $B(1, 7)$ is given by $x(1) + y(7) - (x + 1) - 2(y + 7) - 20 = 0$,which simplifies to $x + 7y - x - 1 - 2y - 14 - 20 = 0$,or $5y = 35$,so $y = 7$.The tangent at $D(4, -2)$ is given by $x(4) + y(-2) - (x + 4) - 2(y - 2) - 20 = 0$,which simplifies to $4x - 2y - x - 4 - 2y + 4 - 20 = 0$,or $3x - 4y = 20$.Solving $y = 7$ and $3x - 4y = 20$ gives $3x - 4(7) = 20$,so $3x = 48$,which means $x = 16$. Thus,$C$ is $(16, 7)$.The quadrilateral $ABCD$ consists of two congruent right-angled triangles $	riangle ABC$ and $	riangle ADC$. The area of $ABCD = 2 	imes 	ext{Area}(	riangle ABC) = 2 	imes (\frac{1}{2} 	imes AB 	imes BC) = AB 	imes BC$.$AB = r = 5$.$BC = \sqrt{(16 - 1)^2 + (7 - 7)^2} = \sqrt{15^2} = 15$.Area of $ABCD = 5 	imes 15 = 75$ square units.

Let $A$ be the center of the circle $x^2 + y^2 - 2x - 4y - 20 = 0$. If $B(1, 7)$ and $D(4, -2)$ are points on the circle,and the tangents at $B$ and $D$ meet at $C$,then the area of the quadrilateral $ABCD$ is:

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