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First calculate the distance between [-sqr(3), 1] and the origin

= sqt ( -sqr(3)^2 + 1^2) = sqt (3+1) = 2

This distance must be equal to the distance between the (s,t) and the origin since it's a circle.

2 = sqt (s^2 + t^2) 4 = s^2 + t^2

Next calculate the slopes. Slope of [-sqr(3), 1] and origin = 1/-sqt(3)
Perpindicular slope must be sqt (3).
Therefore, formula of the line is t = sqt(3)*s + 0

Plug in t = sqt(3)*s inot the above distance equation:
4 = s^2 + [sqt (3)*s]^2
4 = s^2 (1+3)
1 = s

Checking the formulas and eyeballing the graph confirms the answer.

Another approach is to use sin/cos functions.
Given P's coordiantes and radius = 2, you can find out that it makes an angle of 30 degrees with X axis, then Q must make an angle of 60 degrees with X axis, which gives s = 1 (cos 60 = 1/2)