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M09-03

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M09-03  [#permalink]

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New post 16 Sep 2014, 00:38
1
18
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

36% (01:30) correct 64% (02:03) wrong based on 72 sessions

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Re M09-03  [#permalink]

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New post 16 Sep 2014, 00:39
4
7
Official Solution:


THEORY

In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
\((x-a)^2+(y-b)^2=r^2\)

Image

This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled triangle whose other sides are of length x-a and y-b.

If the circle is centered at the origin (0, 0), then the equation simplifies to: \(x^2+y^2=r^2\)

For more check: http://gmatclub.com/forum/math-coordina ... 87652.html

BACK TO THE ORIGINAL QUESTION

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

Curve of \((x - a)^2 + (y - b)^2 = 16\) is a circle centered at the point \(\text{(a, b)}\) and has a radius of \(\sqrt{16}=4\). Now, if \(a\), the x-coordinate of the center, is more than 4 or less than -4 then the radius of the circle, which is 4, won't be enough for curve to intersect with Y axis. So basically the question asks whether \(|a| \gt 4\): if it is, then the answer will be NO: the curve does not intersect with Y axis and if it's not, then the answer will be YES: the curve intersects with Y axis.

(1) \(a^2 + b^2 \gt 16\). Clearly insufficient as \(|a|\) may or may not be more than 4.

(2) \(a = |b| + 5\). As the least value of absolute value (in our case \(|b|\)) is zero then the least value of \(a\) will be 5, so in any case \(|a| \gt 4\), which means that the circle does not intersect the Y axis. Sufficient.


Answer: B
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Re: M09-03  [#permalink]

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New post 01 Oct 2014, 01:59
Hi Bunuel I do not understand this "Now, if a, the x-coordinate of the center, is more than 4 or less than -4 then the radius of the circle, which is 4, won't be enough for curve to intersect with Y axis" Can you explain it for me ? many thanks :D !
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New post 01 Oct 2014, 03:08
this is great.
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Re: M09-03  [#permalink]

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New post 02 Oct 2014, 03:11
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langtuprovn2007 wrote:
Hi Bunuel I do not understand this "Now, if a, the x-coordinate of the center, is more than 4 or less than -4 then the radius of the circle, which is 4, won't be enough for curve to intersect with Y axis" Can you explain it for me ? many thanks :D !


This should be easy if you draw it. If the center is more than 4 units far from y-axis then the radius of 4 won't be enough to reach y-axis:
Attachment:
Untitled.png

>> !!!

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Re: M09-03  [#permalink]

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New post 08 Oct 2014, 13:39
Bunuel wrote:
langtuprovn2007 wrote:
Hi Bunuel I do not understand this "Now, if a, the x-coordinate of the center, is more than 4 or less than -4 then the radius of the circle, which is 4, won't be enough for curve to intersect with Y axis" Can you explain it for me ? many thanks :D !


This should be easy if you draw it. If the center is more than 4 units far from y-axis then the radius of 4 won't be enough to reach y-axis:
Attachment:
Untitled.png


This is great, however i am unable to visualize the case when a co-ordinate is less than -4. Then how exactly will the circle not be able to touch the y axis?
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Re: M09-03  [#permalink]

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New post 08 Oct 2014, 15:01
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earnit wrote:
Bunuel wrote:
langtuprovn2007 wrote:
Hi Bunuel I do not understand this "Now, if a, the x-coordinate of the center, is more than 4 or less than -4 then the radius of the circle, which is 4, won't be enough for curve to intersect with Y axis" Can you explain it for me ? many thanks :D !


This should be easy if you draw it. If the center is more than 4 units far from y-axis then the radius of 4 won't be enough to reach y-axis:
Attachment:
Untitled.png


This is great, however i am unable to visualize the case when a co-ordinate is less than -4. Then how exactly will the circle not be able to touch the y axis?


Check below:
Attachment:
Untitled.png

>> !!!

You do not have the required permissions to view the files attached to this post.


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Re M09-03  [#permalink]

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New post 06 Jan 2015, 09:16
I think this question is good and helpful.
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New post 29 Feb 2016, 23:46
The main theme is whether the circle intersects Y axis or not. So, it depends on 'a'.
First, try to answer without (1) or (2). Can you say the circle intersects Y axis? No. It depends on a. If a is closer enough to Y axis so that 4 unit distance is enough to touch. So, 'a' must be at least within 4 distance or /a/=4 or /a/<4
(1) not sufficient because a2>16-b2. Depending on b, a can be either greater than 4(not touch Y) or less than 4(touch Y). Since it fails to justify a definite answer to the original question, it is not sufficient.
(2) Forget about 1. (2) tells that in any situation or even at the least value of b, 'a' must be at least 5. So, radius 4 is not enough to touch Y axis from 5 unit distance(minimum scenario). So sufficient, and the answer is 'No' to the question. B
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New post 27 May 2016, 22:54
I used algebra approach for this question.

So the circle will intersect with y-axis at point \((0,y_0)\).

Substitute into the question we have: \(a^2 + (y_0 - b)^2 = 16 <-> a^2 - 16 + (y_0 - b)^2 = 0\)
--> so the question is whether this equation have real root \(y_0\) or not.

(1) Lead to: \(a^2 - 16\) can be >0 or <0 --> insufficient
(2) Lead to: \(a^2 - 16 >0 -> a^2 - 16 + (y_0 - b)^2 > 0\): equation has no real root \(y_0\) --> no intersect: sufficient.
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New post 30 Jul 2017, 07:53
I think this is a high-quality question and I agree with explanation. This question really shows why visualization is important in geometry. Awesome Question. Really very thoughtful.
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New post 05 Mar 2019, 08:07
I think this is a high-quality question and I agree with explanation. Very easy question if you can visualize the picture. That's why I love math :)
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Re M09-03   [#permalink] 05 Mar 2019, 08:07
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