One thing that would make this problem immensely easier is to draw it out, so as I'm writing this feel free to draw out what I say.
The first thing to do in this problem is to plot out the circle. The center is at (4,3) and the radius is 2, so we can use this information to find a few other points on the circle. Clockwise from top, they are:
(4,5)
(6,3)
(4,1)
(2,3)
Obviously the circle has more points than these on it but for simplicity we will be focusing mainly on these.
Next we should look at the angle formed by drawing a line from the origin to each of these points and see which is smallest.
Line formed by origin and (4,5) - Fairly large, approximately >45 degrees.
" " " " " (6,3) - Decent size but smaller than the previous line.
" " " " " (4,1) - Very small, smaller than the other lines in fact.
" " " " " (2,3) - Slightly smaller than the first line but bigger than the rest.
We can therefore conclude that the smallest angle with respect to the origin would be formed by the line from the origin to (4,1).
That's nice but how would we find the angle itself? The answer lies in trigonometry. If we were to make the line into a triangle by dropping down a vertical line from the end point and a horizontal line that ended at the same x value as the endpoint, we sould have a triangle. We would also have the measurements of two of the legs of the triangle (4 and 1) and could find the value of the hypotenuse by using the Pythagorean Theorem:
c^2=a^2+b^2
c^2=4^2+1^2
c^2=16+1=17
c=17^0.5 (this is the same as saying the square root of 17)
We also now our equations for sine, cosine, and tangent for each specific angle:
sin(angle x)=length of opposite side/length of hypotenuse
cos(x)=adjacent/hypotenuse
tan(x)=sin(x)/cos(x)=opposite/adjacent
However, because we are trying to find the angle and not the sin/cos/tan of it, we will have to us arcsin/arccos/arctan of the values of the sides of the triangle to find the measure of the angle. Thus, our final step can look like one of the following:
x=arcsin(opposite/hypotenuse)
x=arccos(adjacent/hypotenuse)
x=arctan(opposite/adjacent)
You'll get the same value of x for all 3.
Choosing which angle to find these values of is also an important choice. Because we are trying to find the angle with respect to the x-axis, choosing the angle formed by the x-axis and the line will be the best option. Thus, our final equations will look like this:
x=arcsin(1/(17^0.5))
x=arccos(4/(17^0.5))
x=arctan(1/4)
You'll get the same value of x for all 3.
Hope that helped.
