Find the radius of the circle with equation x^2 + y^2 +8x + 8y + 28 =0 Please help I am so stuck

Question

anonymous · Answer

So what you need to do is complete the square. Do this one at a time. Take the x2 and 8x and add (1/2)(8)^2=16
so you get (x+4)^2
do the same thing with y and get (y+4)^2
Then because you added 16 twice to complete the square twice, you need to subtract 16 twice. This -32 combines with the 28 to make -4
add -4 to other side and get 4
so there you have it the radius is 2
how do i know this?
Because x^2+y^2=r^2
is the general form for a circle.
Now that it is in the form you can pick out r which is on the other side of the equation as 2^2
therefore the radius is 2

anonymous · Answer

$$x^2+y^2+8x+8y+28=0$$
$$x^2+y^2+8x+8y+16+16=-28+16+16$$
$$(x^2+8x+16)+(y^2+8y+16)=4$$
$$(x+4)^2+(y+4)^2=4$$

Now we know that the eq. of circle with centre(h,k) is given by :
$$(x-h)^2+(y-k)^2=r^2$$

Comparing,
h=-4
k=-4
r=2

anonymous · Answer

i have just completed the square .

anonymous · Answer

Thank you guys so much.  It helped a lot.:)

raden · Answer

alternative :
if given an eqution of circle : x^2+y^2+Ax+By+C=0, so its radius = 
r = sqrt (A^2/4 + B^2/4 - C )
r = sqrt (8^2/4 + 8^2/4 - 28 )
r = sqrt (16+16-28)
r = sqrt(4) = 2