Question

The parametric form of the circle $x^2+y^2-4(x+y)=8$ is

• A. $x=2+4\cos \theta ,y=2+4\sin \theta$
• B. $x=2-2\cos \theta ,y=2-2\sin \theta$
• C. $x=-2+2\cos \theta ,y=-2+2\sin \theta$
• D. $x=-2-4\cos \theta ,y=-2-4\sin \theta$

Answer 1

The correct option is A $x=2+4\cos \theta ,y=2+4\sin \theta$

Given equation $x^2+y^2-4(x+y)=8$
$\Rightarrow (x-2{)}^2+(y-2{)}^2=(4{)}^2$
As we know parametric form for the circle
$(x-h{)}^2+(y-k{)}^2=r^2$ is
$x=h+r\cos \theta ,y=k+r\sin \theta$
So, the required parametric form will be
$x=2+4\cos \theta ,y=2+4\sin \theta$