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In the figure above, points P and Q lie on the circle with

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Re: Coordinate System and Geometry [#permalink] New post 19 Nov 2012, 23:59
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Vips0000 wrote:
agarboy wrote:
Can someone explain how to do this problem.. I have no clue how to even start on this.

In the attached picture, from question OA = \sqrt{3}
PA =1
Thus OA = 2


Thus OPA is a 30, 60, 90 triangle. where angle POA is 30.
Also angle POQ is 90, given in question.

therefore, angle QOB = 60
And hence, angle OQB =30

trianlge QOB is also 30, 60, 90 triangle. in which we know OQ =OP =2
Therefore BQ = \sqrt{3}
and OB = 1

We need value of r which is BQ = \sqrt{3}

Can you please explain the red part. It should OP i sppose.
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Re: Coordinate System and Geometry [#permalink] New post 20 Nov 2012, 00:02
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Re: gmatprep geo [#permalink] New post 19 Oct 2013, 08:01
chetanojha wrote:
Check another approach for this problem.


Can we assume that coordinates of the center O are (0,0) here, can we always assume so?
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Re: gmatprep geo [#permalink] New post 20 Oct 2013, 19:35
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obs23 wrote:
chetanojha wrote:
Check another approach for this problem.


Can we assume that coordinates of the center O are (0,0) here, can we always assume so?


From the figure it is clear that O lies at (0, 0). O is the point where x axis and y axis intersect in the figure.
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 10 Dec 2013, 05:34
Bunuel wrote:
zz0vlb wrote:
I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.


In the figure above, points P and Q lie on the circle with center O. What is the value of s?

Yes you are right.

Point Q in I quadrant (1, \sqrt{3});
Point P in II quadrant (-\sqrt{3}, 1);
Point T in III quadrant (-1, -\sqrt{3}) (OT perpendicular to OP);
Point R in IV quadrant (\sqrt{3},-1) (OR perpendicular to OQ).


Is it the case of reflection in Y axis ?
I doubt that but still want to clarify..
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Re: Plane Geometry, Semicircle from GMATPrep [#permalink] New post 10 Dec 2013, 05:40
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ygdrasil24 wrote:
Bunuel wrote:
zz0vlb wrote:
I need someone like Bunuel or walker to look at my statement and tell if this is true.

Since OP and OQ are perpendicular to each other and since OP=OQ=2(radius) , the coordinates of Q are same as P but in REVERSE order, keep the sign +/- based on Quandrant it lies. in this case Q(1,\sqrt{3)}.


In the figure above, points P and Q lie on the circle with center O. What is the value of s?

Yes you are right.

Point Q in I quadrant (1, \sqrt{3});
Point P in II quadrant (-\sqrt{3}, 1);
Point T in III quadrant (-1, -\sqrt{3}) (OT perpendicular to OP);
Point R in IV quadrant (\sqrt{3},-1) (OR perpendicular to OQ).


Is it the case of reflection in Y axis ?
I doubt that but still want to clarify..


No. (-\sqrt{3}, 1) is not a reflection of (1, \sqrt{3}) around the Y-axis. The reflection of (1, \sqrt{3}) around the Y-axis is (-1, \sqrt{3}).
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Re: In the figure above, points P and Q lie on the circle with [#permalink] New post 11 Dec 2013, 11:51
In the figure above, points P and Q lie on the circle with center O. What is the value of s?

This problem took be a bit to conceptualize but I got it as soon as I realized that the graph I was looking at wasn't drawn completely accurately.

If you are to accurately graph the point -√3,1 on a piece of graph paper, you see that the angle it forms with regards to the origin (and the x plane) is less than 45 degrees. Because of this, and the fact that both points lie on the circle, and that the two radii form a 90 degree angle, Q must be higher on the circle than P. Upon looking at this graph, I thought that the x and y coordinates of Q would be the same (except positive) which is, I am sure, the intent of the test makers. The answer is simply the positive reciprocal of point P.

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Re: In the figure above, points P and Q lie on the circle with   [#permalink] 11 Dec 2013, 11:51
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