$\overrightarrow r = \cos \omega t\,\widehat x + \sin \,\omega t\,\widehat y,$

$\therefore$ $\overrightarrow v = - \omega \sin \omega t\widehat x + \omega \cos \omega t\widehat y$

and $\overrightarrow a = - {\omega ^2}\cos \omega t\widehat x - {\omega ^2}\sin \omega t\widehat y$ = $- {\omega ^2}\overrightarrow r$

$\overrightarrow r .\overrightarrow v$ = 0

$\Rightarrow$ $\overrightarrow r \bot \overrightarrow v$

As position vector $\left( {\overrightarrow r } \right)$ is directly away from the origin, so, acceleration ($- {\omega ^2}\overrightarrow r$) is directed towards the origin.