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Circle Question

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Junior | Next Rank: 30 Posts
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by antyagi » Sat Jul 12, 2008 4:23 am
I could not find the solution of the question attached.
This is from GMATPrep
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by sudhir3127 » Sat Jul 12, 2008 5:00 am
OP = OQ ( assume O has coodinates(0,0) )

Distance betwwen two points

(1-0)^2 + (-sqrt3-0)^2 = (s-0)^2 + (t-0)^2
1+3 = s^2+t^2

s^2+t^2 = 4 ---------(1)


Since the lines are perpendicular to each other, product of slopes = -1

(Y2-Y1)/(X2-X1) * (Y2-Y1)/(X2-X1)= -1

[(1-0)/(-sqrt3-0)]*[(t-0)/(s-0)] = -1
t/s = sqrt3
t = s*sqrt3
From (1), s^2 + 3s^2 = 4
s^2 = 1

AS S is in 1st quadrant s=1

hope this helps.
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by antyagi » Sat Jul 12, 2008 5:05 am
Thanks a lot.
Thanks,
Ankur Tyagi
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by rosh26 » Sun Sep 07, 2008 8:40 am
Is there another way to get this solution besides the distance formula? Can we use properties od triangles?
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by rosh26 » Sun Sep 07, 2008 8:41 am
Is there another way to get this solution besides the distance formula? Can we use properties od triangles?
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by Ian Stewart » Sun Sep 07, 2008 9:49 am
If you understand what it means for slopes to be perpendicular, the question can be solved almost immediately. I explained how here:

www.beatthegmat.com/gmat-prep-geometry-t13546.html

Stuart also posted a very clear explanation of this approach in an earlier thread- follow the link from the thread above.

Or, you can notice that you get 30-60-90 triangles if you connect the point P to the x-axis and work from there; that approach is explained in the above thread as well.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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