Answer
$(x-h)^2+(y-k)^2=r^2$
Work Step by Step
Re-arrange the parametric equations as shown below:
$\cos t=\dfrac{x-h}{r}$ and $\sin t=\dfrac{y-k}{r}$
Since, $\sin^2 t+\cos^2 t=1$
$\implies (\dfrac{y-k}{r})^2 + (\dfrac{x-h}{r})^2=1$.
or, $(y-k)^2+(x-h)^2=r^2$
Therefore, $(x-h)^2+(y-k)^2=r^2$
