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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$2 \sqrt{14}$

Vedclass · Answer

(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

Similar Questions

The circle $x^2+y^2+4x-4y+4=0$ touches...

Three circles,each of radius $1$,touch one another externally and lie between two parallel lines. The minimum possible distance between the lines is:

If a circle $S$ with radius $5$ touches the circle $x^2+y^2-6x-4y-12=0$ at $(-1,-1)$,then the length of the tangent from the centre of the circle $S$ to the given circle is

If a circle with center $(-1, 1)$ touches the line $x + 2y + 4 = 0$, then the coordinates of the point of contact are

The number of circles that touch all the three lines $x+y-1=0$,$x-y-1=0$,and $y+1=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module