Finding parametric equations for the curved path of a particle around a half-circle I have a question about parametric equations. So far I've learned how to find the parametric equations for a straight line, I know about replacing $x^2$ and $y^2$ in the equation of the unit circle, but I'm having trouble with this particular problem.
Note that this seems to be a popular problem, there are solutions to it plastered over the Internet, including on StackExchange, but the explanations are not in-depth enough for a newbie of my calibre of the process behind these solutions, that's what I'm struggling with. The question is as follows:
“Find parametric equations for the path of a particle that moves
halfway around the circle $x2 + (y – 1)2 = 4$ counter-clockwise, starting at the point (0 , 3).”
Here is a screenshot of the graph: http://i.imgur.com/Y53jzKQ.png
Could somebody please help me out?

A generic circle of radius $r$ centered at the origin is given parametrically by $\alpha(t) = \langle r\cos(t), r\sin(t) \rangle$ such that $0 \leq t \leq 2\pi$. This is pretty straightforward.  Simply draw a circle centered at the origin and draw a line segment from its center to an arbitrary point on its perimeter.  Call the angle the segment makes with the $x$-axis $t$.  Then, simply think about how $\sin$ and $\cos$ are defined geometrically in terms of a right triangle.  It follows immediately that the $y$-coordinate of your point will be given by $r\sin(t)$, and the $x$-coordinate of your point will be given by $r\cos(t)$.
You can then translate the circle as many units as you'd like in the $x$ or $y$ direction by adding the appropriate number of units to either the $x$ or the $y$ component of the parametrization.  I.e. $\alpha(t) = \langle x + r\cos(t), y + r\sin(t) \rangle$.
Once you have a parameterization for the entire circle centered at the point you'd like, think about what values of $t$ that yield only the half you want.  (Remember that $t$ is the angle between the $x$-axis and a segment from an arbitrary point on your half-circle to the origin).
