## Precalculus (6th Edition) Blitzer

$x=3+6 \cos t$ ; $y=5+6 \sin t$
The standard form of a circle is: $(x-h)^2+(y-k)^2=r^2$ $h=3, k=5; r=6$ Now, in parametric from: $x=h+r \cos t =3+6 \cos t$ and $y=y+k \sin t =5+6 \sin t$ So, $x=3+6 \cos t$ ; $y=5+6 \sin t$