In uniform circular motion, the acceleration vector $\vec{a}$ is centripetal and always points from the particle's position $\vec{r}$ toward the center of the circle $\vec{r}_c = x_c\hat{i} + y_c\hat{j}$.

At time $t_1$, the vector from the particle to the center, $\vec{r}_c - \vec{r}_1$, is parallel to the acceleration, which is in the positive x-direction.

$$ \vec{r}_c - \vec{r}_1 = (x_c - x_1)\hat{i} + (y_c - y_1)\hat{j} $$

For this vector to be in the x-direction, its y-component must be zero.

$$ y_c - y_1 = 0 \implies y_c = y_1 $$

The radius of the circle is the magnitude of this vector, so $R = x_c - x_1$.

The particle's velocity changes from $\vec{v}_1 = v\hat{j}$ at $t_1$ to $\vec{v}_2 = -v\hat{i}$ at $t_2$. Since the motion is uniform, the velocity vector rotates at a constant rate. This change represents a counter-clockwise rotation of $\pi/2$ radians. Therefore, the particle's angular displacement is $\Delta\theta = \pi/2$ over the time interval $\Delta t = t_2 - t_1$.

The arc length $s$ traveled by the particle can be expressed in two ways: $s = v\Delta t$ and $s = R\Delta\theta$. Equating these gives the radius:

$$ v\Delta t = R\Delta\theta $$ $$ v\Delta t = R(\pi/2) \implies R = \frac{2v\Delta t}{\pi} $$

Finally, we find the x-coordinate of the center using the relation derived earlier:

$$ x_c = x_1 + R = x_1 + \frac{2v\Delta t}{\pi} $$