Question 55199
The general form of the equation for a circle with its center at (h, k) and radius of r is:
{{{(x-h)^2+(y-k)^2 = r^2}}}

Let's get your equation into this form, then we can find the center (h, k) and the radius (r).  Starting with the given equation:
{{{x^2+y^2+6x+8 = 0}}} Complete the square in the x-terms by adding 1 to the 8 and 1 to the other side of the equation:
{{{(x^2+6x+9) + y^2 = 1}}} Factor the parentheses.
{{{(x+3)^2 + y^2 = 1}}} Compare with the general form:
{{{(x-h)^2 + (y-k)^2 = r^2}}}

You can see that the center (h, k) is (-3, 0) because h = -3 and k = 0 and the radius (r) is 1 because {{{r^2 = 1}}} and {{{r = sqrt(1)}}}.