If the midpoint of a chord of the circle x^2 | vedclass

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

Question

(D) The equation of the circle is $x^2 + y^2 + x - y - 1 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 1/2$,$f

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The equation of the circle is $x^2 + y^2 + x - y - 1 = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 1/2$,$f = -1/2$,and $c = -1$.The center of the circle is $C(-g, -f) = (-1/2, 1/2)$.The radius $r$ of the circle is $\sqrt{g^2 + f^2 - c} = \sqrt{(1/2)^2 + (-1/2)^2 - (-1)} = \sqrt{1/4 + 1/4 + 1} = \sqrt{3/2}$.Let $M(1, 1)$ be the midpoint of the chord. The distance $d$ from the center $C(-1/2, 1/2)$ to the midpoint $M(1, 1)$ is given by the distance formula:$d = \sqrt{(1 - (-1/2))^2 + (1 - 1/2)^2} = \sqrt{(3/2)^2 + (1/2)^2} = \sqrt{9/4 + 1/4} = \sqrt{10/4} = \sqrt{5/2}$.For a chord to exist,the distance $d$ from the center to the midpoint must be less than the radius $r$ of the circle.Here,$d = \sqrt{5/2} \approx 1.58$ and $r = \sqrt{3/2} \approx 1.22$.Since $d > r$,the point $(1, 1)$ lies outside the circle.Therefore,no such chord exists.

If the midpoint of a chord of the circle $x^2 + y^2 + x - y - 1 = 0$ is $(1, 1)$,then the length of the chord is

Similar Questions

Which of the following is a point on the common chord of the circles ${x^2 + y^2 + 2x - 3y + 6 = 0}$ and ${x^2 + y^2 + x - 8y - 13 = 0}$?

Tangents are drawn from the point $(4, 3)$ to the circle $x^2 + y^2 = 9$. The area of the triangle formed by them and the line joining their points of contact is

The length of the common chord of the two circles $x^2+y^2-4y=0$ and $x^2+y^2-8x-4y+11=0$ is

The point of intersection of the direct common tangents drawn to the circles $(x+11)^2+(y-2)^2=225$ and $(x-11)^2+(y+2)^2=25$ is

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2-3x+y-10=0$ and $x^2+y^2-x+2y-20=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module