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(C) The equation of the circle is ${x^2} + {y^2} = 3$,so the radius $r = \sqrt{3}$ and the center is $C(0, 0)$.Let the chord be $AB$ with length $2$.

Vedclass · Accepted Answer

$\pm \sqrt{10}$

Vedclass · Answer

(C) The equation of the circle is ${x^2} + {y^2} = 3$,so the radius $r = \sqrt{3}$ and the center is $C(0, 0)$.Let the chord be $AB$ with length $2$. The perpendicular distance $d$ from the center $C(0, 0)$ to the chord $x - 2y - k = 0$ bisects the chord.Thus,the distance $d = \sqrt{r^2 - (	ext{half-chord length})^2} = \sqrt{(\sqrt{3})^2 - (1)^2} = \sqrt{3 - 1} = \sqrt{2}$.The perpendicular distance from $(0, 0)$ to $x - 2y - k = 0$ is given by $d = \frac{|0 - 2(0) - k|}{\sqrt{1^2 + (-2)^2}} = \frac{|-k|}{\sqrt{5}}$.Equating the two expressions for $d$: $\frac{|k|}{\sqrt{5}} = \sqrt{2}$.$|k| = \sqrt{10}$,which implies $k = \pm \sqrt{10}$.

If the line $x - 2y = k$ cuts off a chord of length $2$ from the circle ${x^2} + {y^2} = 3$,then $k =$

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