Question

The parametric eqauation of the circle $x^2+y^2-2x-4y-4=0$ is .

• A. $x=1-3\cos \theta ,y=2+3\sin \theta$
• B. $x=1+3\cos \theta ,y=2-3\sin \theta$
• C. $x=1+3\cos \theta ,y=2+3\sin \theta$
• D. $x=1-3\cos \theta ,y=2-3\sin \theta$

Answer 1

The correct option is C $x=1+3\cos \theta ,y=2+3\sin \theta$

We have,
$x^2+y^2-2x-4y-4=0\Rightarrow (x^2-2x+1)+(y^2-4y+4)=9\Rightarrow (x-1{)}^2+(y-2{)}^2=3^2$
Hence the parametric form of the given circle will be :
$x=1+3\cos \theta ,y=2+3\sin \theta$