Question

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

Answer 1

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

Centre of circle $=(3,4)$

Radius of circle $=8$

Therefore,

Corresponding parametric equations are-

$x\left(\theta \right)=h+r\cos \theta =3+8\cos \theta$

$y\left(\theta \right)=k+r\sin \theta =4+8\sin \theta$

where $0\le \theta <2\pi$