If alpha neq -4 and (2, alpha) is the mid-poi | vedclass

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(A) The equation of the circle is $x^2+y^2-4x+8y+6=0$. The center is $(2, -4)$ and the radius is $r = \sqrt{2^2+(-4)^2-6} = \sqrt{4+16-6} = \sqrt{14}$

Vedclass · Accepted Answer

$(-4-\sqrt{14}, -4+\sqrt{14})$

Vedclass · Answer

(A) The equation of the circle is $x^2+y^2-4x+8y+6=0$. The center is $(2, -4)$ and the radius is $r = \sqrt{2^2+(-4)^2-6} = \sqrt{4+16-6} = \sqrt{14}$.Since $(2, \alpha)$ is the mid-point of a chord,it must lie inside the circle. Substituting $(2, \alpha)$ into the circle equation gives $2^2+\alpha^2-4(2)+8\alpha+6 Solving $\alpha^2+8\alpha+2=0$ gives $\alpha = -4 \pm \sqrt{14}$. Thus,$\alpha \in (-4-\sqrt{14}, -4+\sqrt{14})$.The chord passes through $(2, \alpha)$ and is perpendicular to the radius connecting $(2, -4)$ and $(2, \alpha)$. The slope of the radius is undefined (vertical line $x=2$),so the chord is a horizontal line $y=\alpha$.The $y$-intercept of the line $y=\alpha$ is simply $\alpha$. Therefore,the $y$-intercept lies in the interval $(-4-\sqrt{14}, -4+\sqrt{14})$.

If $\alpha \neq -4$ and $(2, \alpha)$ is the mid-point of a chord of the circle $x^2+y^2-4x+8y+6=0$,then the values of the $y$-intercept of the chord lie in the interval

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