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If $\alpha \neq-4$ and $(2, \alpha)$ is the mid-point of a chord of the circle $x^2+y^2-4 x+8 y+6=0$, then the values of the $y$-intercept of the chord lie in the interval
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Verified Answer
The correct answer is:
$(-4-\sqrt{14},-4+\sqrt{14})$
We have,
$(2, \alpha)$ is mid-point of chord of circle
$x^2+y^2-4 x+8 y+6=0$
$\because(2, \alpha)$ inside the circle
$\therefore \quad 4+\alpha^2-8+8 \alpha+6 \leq 0$
$\alpha^2+8 \alpha+2 \leq 0$
$\alpha=\frac{-8 \pm \sqrt{64-8}}{2}$
$\alpha=\frac{-8 \pm 2 \sqrt{14}}{2}=-4 \pm \sqrt{14}$
$\alpha \in(-4-\sqrt{14},-4+\sqrt{14})$
$(2, \alpha)$ is mid-point of chord of circle
$x^2+y^2-4 x+8 y+6=0$
$\because(2, \alpha)$ inside the circle
$\therefore \quad 4+\alpha^2-8+8 \alpha+6 \leq 0$
$\alpha^2+8 \alpha+2 \leq 0$
$\alpha=\frac{-8 \pm \sqrt{64-8}}{2}$
$\alpha=\frac{-8 \pm 2 \sqrt{14}}{2}=-4 \pm \sqrt{14}$
$\alpha \in(-4-\sqrt{14},-4+\sqrt{14})$
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