The minimum distance between the curves x^2 + | vedclass

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(C) The circle equation is $x^2 + 4x + y^2 + 16y + 66 = 0$,which can be written as $(x+2)^2 + (y+8)^2 = 4$. The center is $C(-2, -8)$ and radius $r =

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$4\sqrt{2} - 2 	ext{ units}$

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(C) The circle equation is $x^2 + 4x + y^2 + 16y + 66 = 0$,which can be written as $(x+2)^2 + (y+8)^2 = 4$. The center is $C(-2, -8)$ and radius $r = 2$.The parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$.The minimum distance between the curves lies along the common normal. The normal to the parabola $y^2 = 8x$ at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$. Substituting $a=2$,the normal is $y = -tx + 4t + 2t^3$.Since this normal passes through the center of the circle $(-2, -8)$,we have $-8 = -t(-2) + 4t + 2t^3$,which simplifies to $2t^3 + 6t + 8 = 0$,or $t^3 + 3t + 4 = 0$. By inspection,$t = -1$ is a root.The point on the parabola corresponding to $t = -1$ is $(a(-1)^2, 2a(-1)) = (2, -4)$.The distance from the center $C(-2, -8)$ to the point $P(2, -4)$ is $d = \sqrt{(2 - (-2))^2 + (-4 - (-8))^2} = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}$.The minimum distance between the curves is $d - r = 4\sqrt{2} - 2$.

The minimum distance between the curves $x^2 + y^2 + 4x + 16y + 66 = 0$ and $y^2 = 8x$ is:

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