Answer
The x coordinate of the center of the circular path is $~~x = 8.82~m$
Work Step by Step
The speed of the particle is $v = 3.00~m/s$
At $t_1 = 4.00~s$, the velocity vector is directed in the +y direction and the acceleration vector is in the +x direction. Since the acceleration vector is directed toward the center of the circular path, the particle is moving clockwise.
At $t_2 = 10.0~s$, the particle is moving in the -x direction. Therefore, the particle completed $\frac{3}{4}$ of a circle in $6.00~s$. The particle would complete one full circle in $8.00~s$
We can find the radius:
$v = \frac{2\pi~r}{T}$
$r = \frac{v~T}{2\pi}$
$r = \frac{(3.00~m/s)(8.00~s)}{2\pi}$
$r = 3.82~m$
The center of the circular path is located a distance of $3.82~m$ to the right of the point $(5.00~m, 6.00~m)$
The center of the circular path is located at the point $(8.82~m, 6.00~m)$
The x coordinate of the center of the circular path is $~~x = 8.82~m$
