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Re: In the figure above, points P and Q lie on the circle with [#permalink]
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christoph wrote:


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


A. \(\frac{1}{2}\)

B. \(1\)

C. \(\sqrt{2}\)

D. \(\sqrt{3}\)

E. \(\frac{\sqrt{2}}{2}\)


Attachment:
image.JPG





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Re: In the figure above, points P and Q lie on the circle with [#permalink]
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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Re: In the figure above, points P and Q lie on the circle with [#permalink]
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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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In the figure above, points P and Q lie on the circle with [#permalink]
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]
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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]
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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]
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­Should be no sweat really. All about using the slope:

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