For a particle in uniform circular motion, the acceleration $$\vec{a}$$ at any point $$P(R, \theta)$$ on the circular path of radius $$R$$ is (when $$\theta$$ is measured from the positive $$x$$-axis and $$v$$ is uniform speed):

For a particle in uniform circular motion of radius $$R$$, the position at angle $$\theta$$ from the positive x-axis is:

$$\vec{r} = R\cos\theta\,\hat{i} + R\sin\theta\,\hat{j}$$

In uniform circular motion, the acceleration is centripetal (directed toward the center), i.e., opposite to the position vector from the center:

$$\vec{a} = -\frac{v^2}{R}\hat{r}$$

where $$\hat{r}$$ is the unit vector along the radial direction (outward from center).

The unit radial vector at angle $$\theta$$ is:

$$\hat{r} = \cos\theta\,\hat{i} + \sin\theta\,\hat{j}$$

Therefore:

$$\vec{a} = -\frac{v^2}{R}(\cos\theta\,\hat{i} + \sin\theta\,\hat{j})$$

$$\vec{a} = -\frac{v^2}{R}\cos\theta\,\hat{i} - \frac{v^2}{R}\sin\theta\,\hat{j}$$

Hence, the correct answer is Option C.