Answer
Counterclockwise along the circle $\left(\frac{x-5}{2}\right)^{2}+\left(\frac{y-3}{2}\right)^{2}$ = $1$ from $(3,3)$ to $(7,3)$, the lower half of a circle.
Work Step by Step
$x$ = $5+2\cos{\pi}t$
$\cos{\pi}t$ = $\frac{x-5}{2}$
$y$ = $5+2\sin{\pi}t$
$\sin{\pi}t$ = $\frac{y-3}{2}$
Use the Pythagorean Identity:
$\cos^2 z+\sin^2 z=1$
$\cos^{2}{\pi}t+\sin^{2}{\pi}t$ = $(\frac{x-5}{2})^{2}+(\frac{y-3}{2})^{2}$ = $1$
The motion of the particle takes place on circle centered at $(5,3)$ with a radius $2$. As $t$ goes from $1$ to $2$, the particle starts at the point $(3,3)$ and moves counterclockwise along the circle $\left(\frac{x-5}{2}\right)^{2}+\left(\frac{y-3}{2}\right)^{2}$ = $1$ to $(7,3)$, one half of a circle (the lower part).
