Let the line x+y=1 meet the circle x^2+y^2=4 | vedclass

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(B) The line $AB$ is $x+y=1$. The slope of $AB$ is $-1$. The line perpendicular to $AB$ passing through the origin (since the midpoint of $AB$ lies on

Vedclass · Accepted Answer

$2 \sqrt{14}$

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(B) The line $AB$ is $x+y=1$. The slope of $AB$ is $-1$. The line perpendicular to $AB$ passing through the origin (since the midpoint of $AB$ lies on the line $y=x$) is $y=x$.Solving $y=x$ with $x^2+y^2=4$,we get $2x^2=4$,so $x^2=2$,$x=\pm \sqrt{2}$. Thus,$C=(\sqrt{2}, \sqrt{2})$ and $D=(-\sqrt{2}, -\sqrt{2})$.The length of the chord $CD$ is the diameter of the circle,which is $2r = 2(2) = 4$.The distance of the chord $AB$ from the origin $(0,0)$ is $d = \frac{|0+0-1|}{\sqrt{1^2+1^2}} = \frac{1}{\sqrt{2}}$.The length of the chord $AB$ is $2\sqrt{r^2-d^2} = 2\sqrt{4-\frac{1}{2}} = 2\sqrt{\frac{7}{2}} = \sqrt{14}$.Since $CD$ is perpendicular to $AB$,the area of the quadrilateral $ADBC$ is $\frac{1}{2} 	imes 	ext{diagonal}_1 	imes 	ext{diagonal}_2 = \frac{1}{2} 	imes AB 	imes CD = \frac{1}{2} 	imes \sqrt{14} 	imes 4 = 2\sqrt{14}$.

Let the line $x+y=1$ meet the circle $x^2+y^2=4$ at the points $A$ and $B$. If the line perpendicular to $AB$ and passing through the midpoint of the chord $AB$ intersects the circle at $C$ and $D$,then the area of the quadrilateral $ADBC$ is equal to

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