## Precalculus (6th Edition) Blitzer

$(x-h)^2+(y-k)^2=r^2$
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, $(x-h)^2+(y-k)^2=r^2$