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

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(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

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(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

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