Answer:
[tex]x(t)=8 \cos(t) -3\\\\y(t)=8 \sin(t)-5[/tex]
Step-by-step explanation:
The equation of the circle centered at (-3,-5) in the xy-plane is
[tex](x+3)^2+(y+5)^2=8^2[/tex]
hence in vector parametrization, we have
[tex]x+3=8\cos(t)\\\\y+5=8\sin(t)[/tex]
and so
[tex]x(t)=8\cos(t)-3\\\\y(t)=8\sin(t)-5[/tex]
Moreover, note that
[tex]x(0)=8\cos(0)-3=8-3=5\\\\y(0)=8\sin(0)-5=0-5=-5[/tex]
Otherwise, we should have used a parametrization and we are done.
