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GMAT PREP PS QUESTION 1

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by stop@800 » Sat Sep 20, 2008 11:53 am
A good question, It made me think for a while. :)

Slope of line PO = -1/sqrt(3)
because two points are (0,0) and (-sqrt(3), 1)

Also radius of circle from two points (0,0) and (-sqrt(3), 1) is 2.

equn of circle x^2+y^2 = 4

Line OQ is perpendicular to Op
so slope = sqrt(3)
and equn => y = sqrt(3) * x

Now we have two equations at point Q
s^2 + t^2 = 4
and
t = sqrt(s)

Solve you will get your answer.

I am sure there must be some shortcut also, but as of now I can not think of.


A request: If possible please type the question. Lot of us will be able to save some time. Thanks a lot.
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by sparsh.21 » Sat Sep 20, 2008 1:09 pm
Very good explanation !!!
thanks stop !!
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