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# Does the curve (x - a)^2 + (y - b)^2 = 16

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Senior Manager
Joined: 20 Dec 2004
Posts: 251
Does the curve (x - a)^2 + (y - b)^2 = 16 [#permalink]

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29 Mar 2008, 12:22
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Does the curve $$(x - a)^2 + (y - b)^2 = 16$$ intersect the $$Y$$ axis?

1. $$a^2 + b^2 > 16$$
2. $$a = |b| + 5$$
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Joined: 17 Nov 2007
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29 Mar 2008, 14:09
B

$$(x - a)^2 + (y - b)^2 = 16$$ is a circle with center at (a,b) and radius of 4.

1. $$a^2 + b^2 > 16$$ presents a plane with a hole at the center of coordinates. So, we can choose (a,b) at Y-axis and far away Y- axis. Insufficient.

2. $$a = |b| + 5$$ presents a semi-plane with a>=5. Therefore, any circle with the center at the semi-pane and radius 4 will not intersect Y-axis. Sufficient.
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Senior Manager
Joined: 24 Feb 2008
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29 Mar 2008, 16:32
I don't understand those plane explanations, Walker. If this is a GMAT problem, there has to be an easier explanations.

I paraphrase the question to ask is a in [-4,4], meaning it intersects the Y-axis, or is it not? a is the distance of moving left or right on the X-axis and since we know the radius is 4, a can tell us if the circle intersects the Y-axis.

1. Not sufficient because a can be a number in the interval and outside of it as well.

2. Sufficient because it says a > 5 and that means, a is outside the interval and therefore cannot intersect the Y-axis.
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29 Mar 2008, 21:13
Maybe you are right. I like drawing, because it is fastest way for me
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08 May 2008, 23:24
I found question this sort of tough!! Are such complex problems expected in the real GMAT aor is my coordinate Geo weak?
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08 May 2008, 23:41
Neelesh likes Manhattan Math!
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10 May 2008, 08:40
walker wrote:
B

$$(x - a)^2 + (y - b)^2 = 16$$ is a circle with center at (a,b) and radius of 4.

1. $$a^2 + b^2 > 16$$ presents a plane with a hole at the center of coordinates. So, we can choose (a,b) at Y-axis and far away Y- axis. Insufficient.

2. $$a = |b| + 5$$ presents a semi-plane with a>=5. Therefore, any circle with the center at the semi-pane and radius 4 will not intersect Y-axis. Sufficient.

Walker or anyone: if you could explain how to draw these equations on a plane, it would be very helpful.
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10 May 2008, 11:29
How do you know it is a circle with radius 4 just by seeing the equation and not looking at 1 or 2?
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10 May 2008, 23:13
RyanDe680 wrote:
How do you know it is a circle with radius 4 just by seeing the equation and not looking at 1 or 2?

$$(x - a)^2 + (y - b)^2 = r^2$$ - is a general equation for a circle with center (a,b) and radius of r. So, r=4
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Re: DS: Curve   [#permalink] 10 May 2008, 23:13
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