Question

Obtain the parametric equation of the following circle.
$(x-3{)}^2+(y-4{)}^2=8^2$.

Answer 1

${(x-3)}^2+{(y-4)}^2=8^2$

$r=8$, center$=(3,4)$

Parametric equations are $x=3+8\cos \theta$

$y=4+8\sin \theta$

$\therefore x=3+8\cos \theta ,y=4+8\sin \theta$