Answer
$ x=4+9 \cos t $ ; $ y=6+9 \sin t $
Work Step by Step
The standard form of a circle:
$(x-h)^2+(y-k)^2=r^2$
$ h=4, k=6; r=9$
Now, in parametric from:
$ x=h+r \cos t =4+9 \cos t $
and $ y=y+k \sin t =6+9 \sin t $
So, $ x=4+9 \cos t $ ; $ y=6+9 \sin t $
