Question 1164378
.
<pre>

The given equation is EQUIVALENT to

    {{{x^2 + 6x + y^2}}} = 0


(after dividing both sides by 3).



Next, the last equation you transform using "completing the square"

    {{{(x^2 + 2*3x + 3^2)}}} + {{{y^2}}} = {{{3^2}}}

    {{{(x+3)^2}}} + {{{y^2}}} = {{{3^2}}}.



It describes the circle of the radius 3 centered at the point  (-3,0).
</pre>

Solved, answered and explained.