Question

Parametric equation of the circle $x^2+y^2-6x+8y+16=0$ are

• A. $x=3+3\cos \theta ,y=-4+3\sin \theta$
• B. $x=3+3\cos \theta ,y=-4+3\sin \theta$
• C. $x=-3+3\cos \theta ,y=-4+3\sin \theta$
• D. $x=-3-3\cos \theta ,y=-4-3\sin \theta$

Answer 1

The correct option is B $x=3+3\cos \theta ,y=-4+3\sin \theta$

Given equation
$x^2+y^2-6x+8y+16=0$
$\Rightarrow x^2-6x+9+y^2+8y+16+16-9-16=0$
$\Rightarrow (x-3{)}^2+(y+4{)}^2=3^2$
As we know parametric equation for the circle
$(x-a{)}^2+(y-b{)}^2=r^2$ is
$x=a+r\cos \theta ,y=b+r\sin \theta$
So the required parametric equation will be
$x=3+3\cos \theta ,y=-4+3\sin \theta$