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Solution:
Radius is PO = sqrt(3+1) = 2
PO is perpendicular to QO.
So, (slope of PO) * (slope of QO) = -1.
So, slope of QO is sqrt3.
Or t/s = sqrt3.
Also, QO is the radius.
So, s^2 + t^2 = 4.
Or s^2 + 3s^2 = 4.
Or s^2 = 1. So, s = +1 or -1.
Since Q is in the first quadrant, s is positive.
So, s = 1.
Radius is PO = sqrt(3+1) = 2
PO is perpendicular to QO.
So, (slope of PO) * (slope of QO) = -1.
So, slope of QO is sqrt3.
Or t/s = sqrt3.
Also, QO is the radius.
So, s^2 + t^2 = 4.
Or s^2 + 3s^2 = 4.
Or s^2 = 1. So, s = +1 or -1.
Since Q is in the first quadrant, s is positive.
So, s = 1.
Anurag Mairal, Ph.D., MBA
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GMAT Expert, Admissions and Career Guidance
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Join Our Facebook Groups
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