Circle C is in the xy-plane, what is the area of the circle? : GMAT Data Sufficiency (DS)
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# Circle C is in the xy-plane, what is the area of the circle?

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Circle C is in the xy-plane, what is the area of the circle? [#permalink]

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25 Aug 2010, 03:46
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Circle C is in the xy-plane, what is the area of the circle?

(1) Points (-2, 0) and (0,2) lie on the circle.
(2) The radius of the circle is equal to or less than 2^(1/2).
[Reveal] Spoiler: OA
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25 Aug 2010, 04:29
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jitendra wrote:
Circle C is in the xy-plane, what is the area of the circle?

(1) Points (-2, 0) and (0,2) lie on the circle.
(2) The radius of the circle is equal to or less than 2^(1/2).

$$area=\pi{r^2}$$, so we should find the value of radius.

It would be better if you visualize this problem.

(1) Points (-2, 0) and (0,2) lie on the circle --> two points DO NOT define a circle (three points does), hence we can have numerous circles containing these two points, thus we can not find single numerical value of radius. Not sufficient.

Side note: if you put points (-2, 0) and (0,2) on XY-plane you can see that center of the circle must be on the line $$y=-x$$ (the center of the circle must be equidistant from two pints given).

(2) The radius of the circle is equal to or less than $$\sqrt{2}$$ --> $$r\leq{\sqrt{2}}$$. Clearly insufficient.

(1)+(2) The distance between the 2 points given is $$d=\sqrt{2^2+2^2}=2\sqrt{2}$$, so it's min length of diameter of the circle passing these points (diameter of a circle passing 2 points can not be less than the distance between these 2 points), thus half of $$2\sqrt{2}$$ is min length of the radius of the circle --> $$r\geq{\sqrt{2}}$$ but as from (2) $$r\leq{\sqrt{2}}$$ then $$r=\sqrt{2}$$ --> $$area=\pi{r^2}=2\pi$$. Sufficient.

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25 Aug 2010, 09:03
Good question...As Bunuel said its easier to visualise this and solve than just solving on paper!
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05 May 2011, 21:44
a+b

distance = 2* 2^(1/2) and r <= 2^(1/2)

if r = 1, then diameter = 2 which is less than 2 * 2^(1/2).Not possible.
means r = 2^(1/2).

thus area = pi * [2^(1/2)]^2.

Hence C
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Re: Circle C is in the xy-plane, what is the area of the circle? [#permalink]

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27 Jun 2013, 21:30
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: Circle C is in the xy-plane, what is the area of the circle? [#permalink]

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28 Jun 2013, 21:46
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Adding to Bunnel's explanation, another way to arrive at why r >=\sqrt{2} is because the sum of any two sides of a triangle is greater than or equal to the third side. The triangle under consideration would be the one formed by the points (-2,0) , (0,2) and the center of the circle. Two sides of this triangle are equal to the radii of the circle. Thus, r+r >= distance between points (-2,0) , (0,2) = 2 \sqrt{2}.
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06 Feb 2014, 19:51
Bunuel wrote:
jitendra wrote:

Side note: if you put points (-2, 0) and (0,2) on XY-plane you can see that center of the circle must be on the line $$y=-x$$ (the center of the circle must be equidistant from two pints given).

Can you please elaborate on this point.
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Re: Circle C is in the xy-plane, what is the area of the circle? [#permalink]

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08 Mar 2015, 04:52
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Circle C is in the xy-plane, what is the area of the circle? [#permalink]

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17 Nov 2015, 18:03
neither statement alone is sufficient, yet, when we analyse both statements, we can clearly construct an isosceles triangle with base from 0;2 to 0;-2 or 4 and the angle opposite the base on the center of the circle. since there is only one value for the congruent sides to be equal on the center of the circle while the value of any 2 sides to be greater than the third, technically it is possible to identify the radius. Didn't go further... C
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Re: Circle C is in the xy-plane, what is the area of the circle? [#permalink]

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08 Feb 2016, 16:36
Bunuel wrote:
jitendra wrote:
Circle C is in the xy-plane, what is the area of the circle?

(1) Points (-2, 0) and (0,2) lie on the circle.
(2) The radius of the circle is equal to or less than 2^(1/2).

$$area=\pi{r^2}$$, so we should find the value of radius.

It would be better if you visualize this problem.

(1) Points (-2, 0) and (0,2) lie on the circle --> two points DO NOT define a circle (three points does), hence we can have numerous circles containing these two points, thus we can not find single numerical value of radius. Not sufficient.

Hi:

Please could you help me understand and visualise the highlighted bit.

Thanks
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Re: Circle C is in the xy-plane, what is the area of the circle? [#permalink]

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08 Feb 2016, 17:01
WilDThiNg wrote:
Bunuel wrote:
jitendra wrote:
Circle C is in the xy-plane, what is the area of the circle?

(1) Points (-2, 0) and (0,2) lie on the circle.
(2) The radius of the circle is equal to or less than 2^(1/2).

$$area=\pi{r^2}$$, so we should find the value of radius.

It would be better if you visualize this problem.

(1) Points (-2, 0) and (0,2) lie on the circle --> two points DO NOT define a circle (three points does), hence we can have numerous circles containing these two points, thus we can not find single numerical value of radius. Not sufficient.

Hi:

Please could you help me understand and visualise the highlighted bit.

Thanks

What Bunuel means is that you can not 1 UNIQUE circle by just 2 points. As circle is a planar object, you need to have 3 distinct non collinear points to define a unique circle. Refer to the attached to see that both the circle pass through (-2,0) and (0,2). Clearly the radii of the 2 circles are different ---> giving you 2 different values of the areas. Thus this statement is NOT sufficient.

Attachment:

2016-02-08_19-59-00.jpg [ 32.78 KiB | Viewed 960 times ]

Hope this helps.
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Re: Circle C is in the xy-plane, what is the area of the circle? [#permalink]

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08 Feb 2016, 17:06
Perfect - many thanks!
Re: Circle C is in the xy-plane, what is the area of the circle?   [#permalink] 08 Feb 2016, 17:06
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