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Source: — Problem Solving |

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by Anurag@Gurome » Thu May 12, 2011 8:35 pm
Solution:
Radius is PO = sqrt(3+1) = 2
PO is perpendicular to QO.
So, (slope of PO) * (slope of QO) = -1.
So, slope of QO is sqrt3.
Or t/s = sqrt3.
Also, QO is the radius.
So, s^2 + t^2 = 4.
Or s^2 + 3s^2 = 4.
Or s^2 = 1. So, s = +1 or -1.
Since Q is in the first quadrant, s is positive.
So, s = 1.
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by Bek » Thu May 12, 2011 8:43 pm
Check here:

https://www.beatthegmat.com/gmat-prep-ge ... t4660.html

https://www.beatthegmat.com/what-is-the- ... 27181.html
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by Akansha » Fri May 13, 2011 8:34 pm
Thanks everyone.. That was quite an interesting sum!
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