Question

The parametric form of equation of the circle $x^2+y^2-6x+2y-28=0$ is

• A. $x=-3+\sqrt{38}\cos \theta ,y=-1+\sqrt{38}\sin \theta$
• B. $x=\sqrt{28}\cos \theta ,y=\sqrt{28}\sin \theta$
• C. $x=-3-\sqrt{38}\cos \theta ,y=1+\sqrt{38}\sin \theta$
• D. $x=3+\sqrt{38}\cos \theta ,y=-1+\sqrt{38}\sin \theta$
• E. $x=3+38\cos \theta ,y=-1+38\sin \theta$

Answer 1

Answer: (D)

$x=3+\sqrt{38}\cos \theta ,y=-1+\sqrt{38}\sin \theta$

Given equation is
$x^2+y^2-6x+2y-28=0$
$\Rightarrow (x^2-6x)+(y^2+2y)-28=0$
$\Rightarrow {(x-3)}^2-9+{(y+1)}^2-1-28=0$
$\Rightarrow {(x-3)}^2+{(y+1)}^2={\left(\sqrt{38}\right)}^2$
Hence, parametric equation of the circle will be
$x=3+\sqrt{38}\cos \theta$,
$y=-1+\sqrt{38}\sin \theta$