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In the figure above, points P and Q

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In the figure above, points P and Q [#permalink] New post 24 Dec 2008, 16:45
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Re: Points in a circle [#permalink] New post 25 Dec 2008, 00:37
smarinov wrote:
Attachment:
Q1.JPG


slope of the line OP = 1-0/-SQRT(3)-0 = - 1/SQRT(3)

slope of the line OQ, which is perpendicular to OP = sqrt(3) = t-0/s-0 = t/s

radius=op =
{(-sqrt(3))}^{2} + 1^2=4 =OQ
OQ= 4=
{s*sqrt(3)}^2+ s^2
--> s=1
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Re: Points in a circle   [#permalink] 25 Dec 2008, 00:37
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In the figure above, points P and Q

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