Question

The parametric equation of the circle $x^2+y^2+x+\sqrt{3}y=0$ are

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

Answer 1

The correct option is A $x=\frac{-1}{2}+\cos \theta ,y=\frac{-\sqrt{3}}{2}+\sin \theta$

The parametric equation of circle with radius r and center (a,b) is,

x = a + rcos$\theta$, y = b + rsin$\theta$

The general equation on the other side is,

$(x-a{)}^2+(y-b{)}^2=r^2$

Let's find the radius using equation $x^2+y^2+x+\sqrt{3}y=0$

We can rewrite equation as, $(x+\frac{1}{2}{)}^2+(y+\frac{\sqrt{3}}{2}{)}^2=1$

Comparing this with general equation, we get,

$a=\frac{-1}{2},b=\frac{-\sqrt{3}}{2},r=1$

Hence, the parametric equation of circle is,

x = $\frac{-1}{2}+cos\theta$, y = $\frac{-\sqrt{3}}{2}+sin\theta$