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Does the curve (x-a)^2 + (y-b)^2 = 16 intersect the Y-axis?

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Director
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Joined: 12 Jun 2006
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Does the curve (x-a)^2 + (y-b)^2 = 16 intersect the Y-axis? [#permalink] New post 18 Feb 2007, 01:01
Does the curve (x-a)^2 + (y-b)^2 = 16 intersect the Y-axis?
(1) a^2 = b^2 > 16
(2) a = |b| + 5

With explanations please.
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 [#permalink] New post 18 Feb 2007, 01:38
(D) for me :)

(x-a)^2 + (y-b)^2 = 16 is the formula of a circle with its center at (a,b) and its radius equal to sqrt(16) = 4.

Thus, all we have to know is wether a is far enough on the X axis to not create 1 or 2 intersects between the circrle and the Y axes.

Stat1
a^2 > 16
<=> a <4> 4.

Bingo! in other word, |a| is superior to the radius that is 4, creating no intersects with the Y axes.

SUFF.

Stat2
a = |b| + 5
<b>= 0 (by definition), we have:
a -5 >=5
<a>= 5.

Bingo ! :).... |a| is again greater than the radius 4.

SUFF.
  [#permalink] 18 Feb 2007, 01:38
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Does the curve (x-a)^2 + (y-b)^2 = 16 intersect the Y-axis?

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