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P and Q lie on a circle

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by kartikshah » Sun Jul 15, 2012 12:09 pm
GMAT Prep Practice Test 01
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I chose √3 as my answer but the official correct answer is 1.
Could someone explain the entire problem to me clearly and step by step?
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by eagleeye » Sun Jul 15, 2012 12:31 pm
This question tests two concepts. Properties of a circle and special right angled triangles.
Refer to the image attached:

In triangle PAO, coorinates of P are (-sqrt(3),1). This means that PA is sqrt(3) and OA is 1.
Therefore PAO is a 30-60-90 triangle where AO is 1, PO is sqrt(3) and PO is 2.

Now since PO = OQ = radius of circle = 2, we can find what the co-ordinates of Q are.

We are shown in the question that angle POQ = 90, since we already found that angle POB =60, therefore angle BOQ = 90-60 = 30.

Now triangle BOQ is another 30-60-90 triangle. Since OQ = 2, and BQ is opposite 30 degree angle, BQ = 1. So co-ordinates of Q are (1,sqrt(3). Hence s=1, t= sqrt(3).

Let me know if this helps :)
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by Anurag@Gurome » Sun Jul 15, 2012 6:26 pm
kartikshah wrote:GMAT Prep Practice Test 01
Image

I chose √3 as my answer but the official correct answer is 1.
Could someone explain the entire problem to me clearly and step by step?

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 s = 1.
The correct answer is B.
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by kartikshah » Mon Jul 16, 2012 3:45 am
Anurag@Gurome: Your solution was fine but it didn't work for me. The 30-60-90 rule suggested by eagleeye is something that would click first to me on the real test. Thanks, nevertheless.
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