Question 105703
The equation of a circle centered at (h,k) with radius, R, is given by:
{{{(x-h)^2+(y-k)^2=R^2}}}
{{{(x^2-2hx+h^2)+(y^2-2ky+k^2)=R^2}}}
{{{(x^2-2hx)+(y^2-2ky)=R^2-h^2-k^2}}}
Let's look at your equation and compare terms.
{{{(x^2+8x)+(y^2+6y)-24=0}}}
{{{(x^2+8x)+(y^2+6y)=24}}}
Comparing,
1.{{{-2hx=8x}}}
2.{{{-2ky=6y}}}
3.{{{R^2-h^2-k^2=24}}}
From 1,
1.{{{-2hx=8x}}}
{{{h=-4}}}
From 2, 
2.{{{-2ky=6y}}}
{{{k=-3}}}
From 3,
3.{{{R^2-h^2-k^2=24}}}
{{{R^2-(-4)^2-(-3)^2=24}}}
{{{R^2-16-9=24}}}
{{{R^2=49}}}
{{{R=7}}}
Putting those values back in the first equation, 
{{{(x-h)^2+(y-k)^2=R^2}}}
{{{(x-(-4))^2+(y-(-3))^2=49}}}
{{{highlight((x+4)^2+(y+3)^2=49)}}}
A circle centered at (-4,-3) with a radius of 7. 
Let's verify.
{{{(x+4)^2+(y+3)^2=49}}}
{{{(x^2+8x+16)+(y^2+6y+9)=49}}}
{{{x^2+8x+y^2+6y+25-49=49-49}}}
{{{x^2+8x+y^2+6y-24=0}}}
Back to your original equation. Good answer.