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# A circle with center (1, 0) and radius 2 lies in the

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Director
Joined: 07 Jun 2004
Posts: 604
Location: PA
A circle with center (1, 0) and radius 2 lies in the  [#permalink]

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19 Mar 2011, 06:43
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Difficulty:

85% (hard)

Question Stats:

51% (01:31) correct 49% (02:00) wrong based on 125 sessions

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A circle with center (1, 0) and radius 2 lies in the coordinate plane . If point (x, y) lies on the circle is (x, y) in quadrant 2?

(1) | y | > 1
(2) | x | > 1

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19 Mar 2011, 20:46
3
2
rxs0005 wrote:
A circle with center (1, 0) and radius 2 lies in the coordinate plane . If point (x, y) lies on the circle is (x, y) in quadrant 2?

(1) | y | > 1
(2) | x | > 1

As fluke pointed out, it is a good idea to draw a diagram here. In fact, in most co-ordinate geometry questions, a diagram is almost a must.

Ques: Is (x, y) in quadrant II?
Attachment:

Ques2.jpg [ 8.05 KiB | Viewed 2359 times ]

Stmnt 1: | y | > 1
Look at the red lines above. Here | y | > 1 on the y axis. So (x,y) can be any point on the green region. It may or may not lie in the second quadrant. Not sufficient.

Stmnt 2: | x | > 1
Look at the blue line above. Here | x | > 1 on the x axis. So (x,y) can be any point on the orange line. None of these points lie in the second quadrant. Sufficient.

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##### General Discussion
Intern
Joined: 21 Aug 2010
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19 Mar 2011, 08:12
1
let us consider the 1st statement. y could be y>1 or y<-1. Since we do not know anything about x, the point (x,y) may lie in any of the quadrants. Not suff.
St. 2. If |x|>1, then x>1 or x<-1. But x<-1 is impossible since it lies outside the circle. So we are left with x>1. Is a point (x,y) has x>1, it definitely cannot lie on the second quadrant, but only in the first or fourth. Suff.
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19 Mar 2011, 14:48
1
rxs0005 wrote:
A circle with center (1, 0) and radius 2 lies in the coordinate plane . If point (x, y) lies on the circle is (x, y) in quadrant 2?

(1) | y | > 1
(2) | x | > 1

If you draw a circle, you can solve it quickly.

We can see that for the 2nd quadrant, the (x,y) coordinate must be between $$(0,\sqrt{3})$$ and $$(-1,0)$$.

As we can clearly see that, if the radius is 2 and center is at (1,0); the extreme value of x for 2nd quadrant will be 1-2=-1, where y will be 0.

We know the extreme value for x; i.e. -1

Let's find out the extreme value for y; y will be maximum when x=0;

We can find y where x=0 by Pythagoras theorem;

Distance between origin and the center is 1.
Distance between center and the intersection of circle and y-axis is 2(radius).

Thus, distance between origin and extreme value of y in 2nd quadrant = $$\sqrt{2^2-1^2}=\sqrt{3}\approx 1.73$$

Thus we know the extreme(maximum) value for both x and y in the 2nd quadrant.

$$x<=-1 & y<=1.73$$

1. |y| > 1

Thus y can be 1.2 or 1.8.
1.2<1.73
1.8>1.73

(x,y) may or may not be in the 2nd quadrant
Not Sufficient.

2. |x| > 1

As we know that the maximum value of x in 2nd quadrant is 1. (x,y) can't be in the 2nd quadrant because |x|>1.

Sufficient.

Ans: "B"
*****************

Please take a look at the image to get a visual clarification as well.
Attachments

quadrant_for_point_x_y_on_circle.PNG [ 9.87 KiB | Viewed 2408 times ]

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19 Mar 2011, 17:15
Thanks Fluke. Great explanation.
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
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19 Mar 2011, 20:08
Interesting! Consider this we move the center to origin then both statements are insufficient. But not now.

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Re: A circle with center (1, 0) and radius 2 lies in the  [#permalink]

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28 Dec 2013, 07:08
rxs0005 wrote:
A circle with center (1, 0) and radius 2 lies in the coordinate plane . If point (x, y) lies on the circle is (x, y) in quadrant 2?

(1) | y | > 1
(2) | x | > 1

Statement 1

Statement 2

| x | > 1 means either x>1 or x<-1

Since circle has radius 2 and centered at (1,0) then x cannot be less than -1.

Therefore, x>1

So it CANNOT be on the second quadrant

Hope it helps!
Cheers!
J
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Joined: 09 Sep 2013
Posts: 8101
Re: A circle with center (1, 0) and radius 2 lies in the  [#permalink]

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25 Sep 2017, 10:56
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