Find the centre and radius of the circle:x^2 + y^2 + 8x + 6y= 0

We know that the equationof a circle with centre (h,k) and
radius ris given by:

(x-h)^2+(y-k)^2 =
r^2......(1)

We now recast x^2+y^2+8x+6y = 0 in the form at (1) and
equate the like terms:

X^2+8x = (x+4)^2
-4^2.

y^2+6y = (y+4)^2 -3^2

Therefore
x^2+y^2+8x+6y = (x+4)^2+(y+3)^2 -4^2-3^2.

Therefore the equation
x^2+y^2+8x+6y= 0 is the same as:

(x+4)^2+(y+3)^2 -25 = 0.
Or

(x+4)^2+(y+3)^2 = 25 =
5^2.

{x-(-4)^2}^2 + {y- (-3)}^2 =
5^2.....(2).

Since eq (1) and eq(2) are same we can equate like
terms:

{x-h)^2 = {x-(-4)}^2 gives h =
-4.

Similarly, (y-k)^2 = {y-(-3)}^2  gives k=
-3.

r^2= 5^2 gives  r= 5.

Therefore 
the centre of the circle = (-4,-3) and the radius of the circle = 5.