In the figure shown, point \(P(-\sqrt{3}, 1)\) and \(Q(s, t)\) lie on the circle with center O. What is the value of \(s\)?
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Correct answer: B
As shown in the figure, OP and OQ are perpendicular radii of the circle with center O at the origin.<br><br>
The slope of line OP is \(-\frac{1}{\sqrt{3}}\).<br><br>
Since OP ⊥ OQ, their slopes are negative reciprocals. The slope of OQ is \(\frac{t}{s}\), so:<br>
\[\frac{t}{s} \times \left(-\frac{1}{\sqrt{3}}\right) = -1\]<br>
\[t = \sqrt{3}s\]<br><br>
Since OP is a radius: \(OP = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2\)<br><br>
Since OQ = OP = 2: \(t^2 + s^2 = 4\)<br><br>
Substituting \(t = \sqrt{3}s\):<br>
\[(\sqrt{3}s)^2 + s^2 = 4\]<br>
\[3s^2 + s^2 = 4\]<br>
\[4s^2 = 4\]<br>
\[s^2 = 1\]<br>
\[s = \pm 1\]<br><br>
Since point Q is shown in the first quadrant in the figure (where both coordinates are positive), \(s = 1\).<br><br>
The answer is B.