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Points P and Q lie on a circle

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Points P and Q lie on a circle

by HeyArnold » Wed Sep 28, 2011 5:50 pm
https://postimage.org/image/8okizrdvo/

What got me on this problem--and I think will get me on any similar problem is that both lines start from (0,0) which is the center point of the circle.

If this is the case, from center point to the edge is the radius right? So I would assume OP = OQ

Then, since <POQ is a 90 degree angle, OP and OQ are same length and same angle, so P and Q must be same distance from 0.

Please let me know what I'm missing so I don't make this mistake again!
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by Anurag@Gurome » Wed Sep 28, 2011 5:51 pm
HeyArnold wrote:https://postimage.org/image/8okizrdvo/

What got me on this problem--and I think will get me on any similar problem is that both lines start from (0,0) which is the center point of the circle.

If this is the case, from center point to the edge is the radius right? So I would assume OP = OQ

Then, since <POQ is a 90 degree angle, OP and OQ are same length and same angle, so P and Q must be same distance from 0.

Please let me know what I'm missing so I don't make this mistake again!

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 user123321 » Wed Sep 28, 2011 6:10 pm
alternatively,

slope of the line formed by origin and (-√3,1) is -1/√3
tanx = -1/√3 =>x = 150 degrees
from diagram the resulting line should have 90 degrees less => 150-90=60 degrees
tan60 = √3 = √3/1
by observation if we assume the point to be(1,√3) then it is same distance as other point from origin so should be lying on circle = > s = 1

user123321
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