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

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Bunuel
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Re: In the figure above, points P and Q lie on the circle with [#permalink] New post   01 Apr 2018, 02:31
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ktjorge wrote:
I think I understand how to solve the problem now, but I am having trouble understanding one assumption that people are making in many of the solutions: that O is at the origin. When I tried to solve this problem in the practice test I got to the point where I knew the radius was 2 and that line QO had a slope that was the negative reciprocal of line PO (since they were perpendicular bisectors), but I couldn't solve for slope m of line QO because I was stuck with 1=-sqrt(3)*m+b.

Can we assume that O is at the origin even though it isn't stated in the question, just because it looks like it is? Why can we assume that and not that the y-axis is equidistant from points P and Q (the diagram shows the y-axis passing directly in the middle of the 90 degree angle shown)?


O is the origin because it's on the intersection of x and y axis. Also, if we did not know that, we would not be able to solve the question and since it's a PS questions, we should be able to.

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Re: In the figure above, points P and Q lie on the circle with [#permalink] New post   29 Dec 2019, 22:31
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altairahmad wrote:
1 goofy quetion please chetan2u Bunuel Gladiator59 VeritasKarishma

Since we can calculate that PQ (distance) is 2 root 2 and we know that X coordinate of PO is - root 3 then why isnt x coordinate of QO simply the difference between the two values ?


I am not sure why you think that PQ is 2*sqrt(2). Also, PQ is not a line parallel to x axis.
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Re: In the figure above, points P and Q lie on the circle with [#permalink] New post   08 Apr 2021, 02:51
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Solution:

O is the center.

OP and OQ are perpendicular to each other with OP=OQ (radius)

Thus in the other quadrant (Quadrant I), the coordinates for P would be same as Q but in the reversed order with signs according to the quadrant
=>(1, sqrt3)

=> s =1 (option b)

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Re: In the figure above, points P and Q lie on the circle with [#permalink] New post   10 Dec 2013, 06:34
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: In the figure above, points P and Q lie on the circle with [#permalink] New post   10 Dec 2013, 06:40
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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   31 Mar 2018, 18:33
I think I understand how to solve the problem now, but I am having trouble understanding one assumption that people are making in many of the solutions: that O is at the origin. When I tried to solve this problem in the practice test I got to the point where I knew the radius was 2 and that line QO had a slope that was the negative reciprocal of line PO (since they were perpendicular bisectors), but I couldn't solve for slope m of line QO because I was stuck with 1=-sqrt(3)*m+b.

Can we assume that O is at the origin even though it isn't stated in the question, just because it looks like it is? Why can we assume that and not that the y-axis is equidistant from points P and Q (the diagram shows the y-axis passing directly in the middle of the 90 degree angle shown)?
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Re: In the figure above, points P and Q lie on the circle with [#permalink] New post   27 Dec 2019, 05:07
1 goofy quetion please chetan2u Bunuel Gladiator59 VeritasKarishma

Since we can calculate that PQ (distance) is 2 root 2 and we know that X coordinate of PO is - root 3 then why isnt x coordinate of QO simply the difference between the two values ?
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In the figure above, points P and Q lie on the circle with [#permalink] New post   13 Aug 2020, 12:16
VeritasKarishma wrote:
altairahmad wrote:
1 goofy quetion please chetan2u Bunuel Gladiator59 VeritasKarishma

Since we can calculate that PQ (distance) is 2 root 2 and we know that X coordinate of PO is - root 3 then why isnt x coordinate of QO simply the difference between the two values ?


I am not sure why you think that PQ is 2*sqrt(2). Also, PQ is not a line parallel to x axis.


Hi VeritasKarishma : PQ is 2*\(\sqrt{2}\) because triangle OPQ is an isosceles triangle given 2 sides are equal with a 90 degree in between.

Hence triangle OPQ is a 45-45-90 . Hence the base is 2*\(\sqrt{2}\)

If PQ would have been a straight line parallel to the x axis (I don't think PQ is parallel to the X axis, but if so, could we have done this to find the value of s ?)

PQ = 2*\(\sqrt{2}\) and distance of PZ is \(\sqrt{3}\), then ZQ should be 2*\(\sqrt{2}\) - \(\sqrt{3}\)

(Z is not shown but z is the point where line PQ intersects with the Y axis)

Thoughts ?
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Re: In the figure above, points P and Q lie on the circle with [#permalink] New post   16 Aug 2020, 23:58
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jabhatta@umail.iu.edu wrote:
VeritasKarishma wrote:
altairahmad wrote:
1 goofy quetion please chetan2u Bunuel Gladiator59 VeritasKarishma

Since we can calculate that PQ (distance) is 2 root 2 and we know that X coordinate of PO is - root 3 then why isnt x coordinate of QO simply the difference between the two values ?


I am not sure why you think that PQ is 2*sqrt(2). Also, PQ is not a line parallel to x axis.


Hi VeritasKarishma : PQ is 2*\(\sqrt{2}\) because triangle OPQ is an isosceles triangle given 2 sides are equal with a 90 degree in between.

Hence triangle OPQ is a 45-45-90 . Hence the base is 2*\(\sqrt{2}\)

If PQ would have been a straight line parallel to the x axis (I don't think PQ is parallel to the X axis, but if so, could we have done this to find the value of s ?)

PQ = 2*\(\sqrt{2}\) and distance of PZ is \(\sqrt{3}\), then ZQ should be 2*\(\sqrt{2}\) - \(\sqrt{3}\)

(Z is not shown but z is the point where line PQ intersects with the Y axis)

Thoughts ?


Yes, in that case you could have done this. But do not go by the figure given. It could be drawn to misdirect you. Just because PQ looks parallel to X axis, it doesn't mean it is. Redraw it depending on the data you have and using extreme cases.
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Re: In the figure above, points P and Q lie on the circle with [#permalink] New post   22 Jul 2023, 16:27
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