Drop a perpendicular from point to negative x axis. Let that point be M

The radius of the circle is 2 units.

Angle POQ is 90 degrees. Angle POM is 30 degrees (sides of right triangle with 30 - 60 - 90 degree angles, side opposite side 1 will be 30 degreees)
Which gives angle QOX to be 60 degrees.

Now the radius of the circle is 2. (sides of right triangle with 30 - 60 - 90 degree angles).

Therefore s = 1 and t = sqrt(3)

This need not be a simple solution. I welcome any other simple solution.

Few things we should know to use this solution:
1. Equation of circle: x^2 + y^2 = radius ^ 2
2. Equation of line: y = mx + c, here m is the slope, and c is the y intercept
3. If two lines are perpendicular to each other, then the products of their slopes is -1.

Find the slope of the line with coordinates: (0,0) and (-sqrt(3), 1) =>
-1/sqrt(3).

So, the slope of the line with coordinates: (0,0) and (s, t) =>
sqrt(3).

The equation of the line with coordinates: (0,0) and (s, t) =>
y=sqrt(3)x, y intercept is 0 (the origin).

We need to know the point where this line intersect with the circle. Before this step, find out about the radius of the circle, which is "2".

The equation of this circle is x^2+y^2=4. In this equation, substitute "y=sqrt(3)x", you will get x^2 + 3x^2=4.

4x^2=4 => x^2 = 1. x can be -1 or +1. From this diagram, we can tell x must be +1, and that is the value of s.

If the question asks about "y", then substitute the value "1" for the var x in the equation y = sqrt(3)x.

Regards.