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GMAT Prep - Geometry

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Source: — Problem Solving |

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by Anurag@Gurome » Fri May 18, 2012 11:02 pm
krishna239455 wrote:Image

Point P and Q lies on the same circle with center at (0, 0).
Thus, (s² + t²) = (-√3)² + 1² = 3 + 1 = 4

Again line segments OP and OQ are perpendicular.
Thus (slope of OP)*(slope of OQ) = -1

Slope of OP = 1/(-√3) = -(1/√3)
=> Slope of OQ = (t - 0)/(s - 0) = t/s = (-1)/(-1/√3) = √3
=> t = √3s

Thus, (s² + (√3s)²) = 4
=> (s² + 3s²) = 4
=> s² = 1
=> s = ±1

As point Q lies in the first quadrant, so s = 1.
The correct answer is B.
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by krishna239455 » Fri May 18, 2012 11:31 pm
Hi Anurag

Thanks for the answer:

I tried to solve it by applying the following rule but coul not suceed. Pls Tell me where I am going wrong.

1) This rule is same as applied by you that the distance between Q and origin should be 2.

2) since it makes a right angled isoceles triangle (two equal radii and a right angle) the hypotenuese i.e PQ = 2 sqr root 2..

But this logic does not help. I dont know why? Request you to explain.
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