## Calculus (3rd Edition)

The curve is a circle of radius $3$ and the center is $(4,5)$.
In the previous exercise, the center of the circle is at the origin. But when $x=4+3\cos t$ and $y=5+3\sin t$, then we have $$(x-4)^2+(y-5)^2=9\cos^2t+9\sin^2t=9.$$ So the curve is a circle of radius $3$ and the center is $(4,5)$.