Question 203863
First get your equation into the standard form for a circle whose center is at (h, k) and whose radius is r.:
{{{(x-h)^2+(y-k)^2 = r^2}}}
{{{x^2+y^2+6x = 0}}} Group the variables as shown below:
{{{(x^2+6x)+y^2 = 0}}} Now "complete-the-square" in the x-variable by adding the square of half the x-coefficient to both sides of the equation. {{{(6/2)^2 = 9}}}
{{{(x^2+6x+9)+y^2 = 0+9}}} Factor the trinomial in x.
{{{(x+3)^2+y^2 = 9}}} Compare this with the standard form for a circle:
{{{(x-h)^2+(y-k)^2 = r^2}}}
The center of the circle is at (-3, 0) and the radius is 3.
It's probably easier to graph the circle then pick out the appropriate points inside and outside of the circle.
To graph the circle, you will need to solve your equation for y (you will get two solutions) then graph both solutions.
{{{x^2+y^2+6x = 0}}} Subtract {{{x^2}}} and {{{6x}}} from both sides.
{{{y^2 = -x^2-6x}}} Take the square root of both sides.
{{{y = sqrt(-x^2-6x)}}} and {{{y = -sqrt(-x^2-6x)}}}
Now we'll graph these two solutions on the same grid:
{{{graph(400,400,-8,8,-8,8,sqrt(-x^2-6x),-sqrt(-x^2-6x))}}}