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DS: Curve (m09q03)

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Re: DS: Curve (m09q03) [#permalink] New post 22 Dec 2011, 10:31
No argument regarding the answer, but I do have a bone to pick with this question. Is this really a suitable GMAT preparation question? The Math Review from the OG does not indicate that test-takers are expected to know the equation of a circle in the coordinate plane and the properties of such an equation. And I have yet to come across any official questions testing this type of knowledge. So while the question is interesting and difficult, I feel like it is more of a general interest type of question that a math-lover would tackle in his/her free time rather than a question reflective of the type that would show up on the GMAT... any opinions on this? Have any test-takers seen an "equation of a circle" question on the GMAT before? :?:
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Re: DS: Curve (m09q03) [#permalink] New post 22 Dec 2011, 10:42
The OG 12 does speak of a Circle equation...and it is very much expected from a candidate to know the equation of basic curves including lines, circles, parabolas etc..
This is not a tough problem if you know that you are working with a circle...and can be expected if you are doing well to very well on your test..
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Re: DS: Curve (m09q03) [#permalink] New post 23 Dec 2011, 02:25
Thanks for the reply! Do you think you could provide me with the page number(s) on which this information is mentioned? As far as I can tell, the OG12 Math Review portion on "Circles" only mentions area, circumference, areas of a sector, arc length and inscribed polygons. And the portion on "Coordinate Geometry" only goes up to curves of a parabolic nature. And, I am also unable to find any mention of the equation of a circle in the MGMAT Study Guides that I am using... It is upsetting to know that this topic has been overlooked in them. :(

And yes, I know the question is not difficult once you recognize it to be the equation of a circle. I was just surprised to see such a question come up because in my preparation I had not yet come across this type of question, which explicitly requires you to know the equation of a circle. Usually a question of this nature would be solvable by other (algebraic) means. In this case, without knowledge of the fact that this is the equation of a circle, one would be unable to answer this question (or so I think).
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Re: DS: Curve (m09q03) [#permalink] New post 07 Jan 2012, 17:05
...I hate geometry
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Re: DS: Curve (m09q03) [#permalink] New post 04 Sep 2012, 04:57
bigfernhead wrote:
Does the curve (x - a)^2 + (y - b)^2 = 16 intersect the Y axis?

1. a^2 + b^2 > 16
2. a = |b| + 5

I have no idea how to tackle this... please explain. Thanks.

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You don't need to know geometry, just a little bit of algebra.

The curve (x - a)^2 + (y - b)^2 = 16 intersects the Y axis if for x=0, there is at least one y satisfying the given equation, which becomes a^2+(y-b)^2=16 or (y-b)^2=16-a^2. It is obvious that this equation has at least one solution if 16-a^2\geq{0} or a^2\leq{16}.
Therefore, the stem question can be replaced by "Is a^2\leq{16}?"

(1) Obviously not sufficient.
(2) Since a = |b| + 5\geq{5} , \, a^2\geq{25}>16.
The answer is a definite NO.
Sufficient.

Answer B.
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Re: DS: Curve (m09q03) [#permalink] New post 04 Sep 2012, 22:42
Jdam wrote:
...I hate geometry


What about algebra? See my previous post ds-curve-m09q03-74593-20.html#p1118582
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Re: DS: Curve (m09q03) [#permalink] New post 08 Sep 2012, 04:19
bblast wrote:
+1 fluke

my brains' light finally flickered after ur comment. got it now. :-D


I agree to B.

(x-a)^2 + (y-b)^2 = 16 is a circle with radius 16 and centre (a,b).

If the circle touches y axis, then x = 0 => a^2+ (y-b)^2 = 16
=> (y-b)^2 = 16 - a^2
=> since (y-b)^2 is always positive 16 - a^2 >= 0

=>a^2 <= 16
=> -4 <= a <= 4
a) a ^ 2 + b^ 2 = 16 ( We do not know the value of a or b)
b) a = |b|+4 (|b| is always positive so a >=4)

According to above deduction, the circle touches the y axiz only if -4 <= a <= 4. But Option 2 Says that a >= 4 So we clearly state that Answer Choice is : B
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Re: DS: Curve (m09q03) [#permalink] New post 08 Sep 2012, 04:39
archana1729 wrote:
bblast wrote:
+1 fluke

my brains' light finally flickered after ur comment. got it now. :-D


I agree to B.

(x-a)^2 + (y-b)^2 = 16 is a circle with radius 16 and centre (a,b).

If the circle touches y axis, then x = 0 => a^2+ (y-b)^2 = 16
=> (y-b)^2 = 16 - a^2
=> since (y-b)^2 is always positive 16 - a^2 >= 0

=>a^2 <= 16
=> -4 <= a <= 4
a) a ^ 2 + b^ 2 = 16 ( We do not know the value of a or b)
b) a = |b|+4 (|b| is always positive so a >=4)

According to above deduction, the circle touches the y axiz only if -4 <= a <= 4. But Option 2 Says that a >= 4 So we clearly state that Answer Choice is : B


From the above it is clear that all you have to know is what does it mean a curve given by an equation intersects the Y axis.
You don't have to worry about the type of the curve the equation represents, although it is nice to have the visual interpretation.
The question finally is an algebra question, concerned with the existence of solution for a particular algebraic equation.
Next time, they can give you almost any curve, you will be able to handle the issue of intersection with an axis easily, if the algebra is manageable.
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Re: DS: Curve (m09q03) [#permalink] New post 07 Jan 2013, 04:38
Hi i got statement 1 but as far as (2) is concerned a= !b!+5 then this can be broken as ---> a+b=5 & a-b=5 .
Why we consider 'b" = 0 , we can consider 4+1 =5(4,1) & 5-0= 5 (5,0)...please clarify my doubt
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Re: DS: Curve (m09q03) [#permalink] New post 07 Jan 2013, 05:03
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archit wrote:
Hi i got statement 1 but as far as (2) is concerned a= !b!+5 then this can be broken as ---> a+b=5 & a-b=5 .
Why we consider 'b" = 0 , we can consider 4+1 =5(4,1) & 5-0= 5 (5,0)...please clarify my doubt


We are not saying that b=0. From the stem the question became: is |a|>4? Check here: ds-curve-m09q03-74593.html#p839873

(2) says that a=|b|+5. Now, since the least value of |b| is 0, then the lest value of a is 5, so more than 4. Which means that in any case |a|>4.

Hope it's clear.
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Re: DS: Curve (m09q03) [#permalink] New post 03 Sep 2013, 05:19
GMAT Tiger is WRONG. Answer is B.

All you're doing is looking at shifting a circle with radius of 4, centered at the origin, either up/down or left/right.

A vertical shift does not affect the answer (whether the circle intersects the Y axis), but a horizontal shift will affect the answer. If we shift the entire circle to the right or left by more than 4 units, then it will not intersect the Y axis.

(1) give you nothing, a and b could be anything.

(2) tells you that 'a' is greater than 5, so the circle has been shifted by 5 units, thereby causing it to not intersect the Y axis (as radius is only 4).

So answer is B, (2) gives you everything you need to know.

You can see what the author of the question is trying to get at, that helps a lot when trying to solve these problems.
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Re: DS: Curve (m09q03) [#permalink] New post 03 Sep 2013, 17:37
the given equation is an circle center (a,b) radius 4
so the circle will intersect y axis if radius greater than the distance of center from y axis,i.e the x co-ordinate of the center 'a'
hence b is sufficient
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Re: DS: Curve (m09q03) [#permalink] New post 04 Sep 2013, 08:41
The answer is it can be answered by using any one of the two statements.

From the equation of the circle given it is clear that the center lies in 1st quadrant.
Let us consider the area enclosed by the (a,b) with the x axis and y axis . x-intercept is a and y intercept is b.
So in order for the circle to touch any of the x ,y or both(origin) (a^2+b^2)=16.so that circle touches the axes.
But it is clearly given in the first statement it is greater than 16.So the circle do not touch y axis.

In statement 2 the x coordinate is greater than 5( minimum value). So the circle do not touch the yaxis.

So the question can be answered by either of the statements :-D
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Re: DS: Curve (m09q03) [#permalink] New post 04 Sep 2013, 09:22
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jayavignesh wrote:
The answer is it can be answered by using any one of the two statements.

From the equation of the circle given it is clear that the center lies in 1st quadrant.
Let us consider the area enclosed by the (a,b) with the x axis and y axis . x-intercept is a and y intercept is b.
So in order for the circle to touch any of the x ,y or both(origin) (a^2+b^2)=16.so that circle touches the axes.
But it is clearly given in the first statement it is greater than 16.So the circle do not touch y axis.

In statement 2 the x coordinate is greater than 5( minimum value). So the circle do not touch the yaxis.

So the question can be answered by either of the statements :-D


Note that the correct answer is B, not D. Check here: ds-curve-m09q03-74593.html#p839873

Hope it helps.
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Re: DS: Curve (m09q03) [#permalink] New post 05 Sep 2013, 22:37
jayavignesh wrote:
The answer is it can be answered by using any one of the two statements.

From the equation of the circle given it is clear that the center lies in 1st quadrant.
Let us consider the area enclosed by the (a,b) with the x axis and y axis . x-intercept is a and y intercept is b.
So in order for the circle to touch any of the x ,y or both(origin) (a^2+b^2)=16.so that circle touches the axes.
But it is clearly given in the first statement it is greater than 16.So the circle do not touch y axis.

In statement 2 the x coordinate is greater than 5( minimum value). So the circle do not touch the yaxis.

So the question can be answered by either of the statements :-D

How can you say that the center of the circle lies in first quadrant? It depends on the sign of a and b.
One way to look at stmt 1 is, as a^2+b^2>16 , the distance of (a,b) is more than 4 from origin. With such a and b there can be numerous circles, some of them will cut the y axis and some will not. Hence not sufficient.
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Re: DS: Curve (m09q03) [#permalink] New post 19 Aug 2014, 15:14
bigfernhead wrote:
Does the curve (x - a)^2 + (y - b)^2 = 16 intersect the Y axis?

1. a^2 + b^2 > 16
2. a = |b| + 5

I have no idea how to tackle this... please explain. Thanks.

[Reveal] Spoiler: OA
B

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the equation is a classic circle with center at (a, b) and radius 4.

Statement 1 gives a relationship between a and b, which alone is insufficient.

Statement 2 gives a relationship between a and b.

a = |b| + 5
Taking a as 0, |b| = -5 or <0, which is impossible since |b| has to be positive.
Least value of |b| could only be 0, hence least value of a is 5.

So any circle with radius 4 having its center on the vertical line with X = 5 is never going to intersect the y-axis. Sufficient

IMO (B)
Re: DS: Curve (m09q03)   [#permalink] 19 Aug 2014, 15:14
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