# pre-calculus

posted by on .

Determine the center and the radius of the following circle:

x^2+y^2+8x-2y=8

so I figured I'd use completing the square after putting it in standard form:

(x^2+8x)+(y^2-2y)=8

I'm not sure where to go from here.

• pre-calculus - ,

but you didn't "complete" the square

(x^2+8x)+(y^2-2y)=8
(x^2+8x + 16)+(y^2-2y + 1)= 8 + 16 + 1
(x+4)^2 + (y-1)^2 = 25

Now we have it in standard form.
Can you take it from there ?

• pre-calculus - ,

Ok. Do I take the square root of the whole thing now?

getting x+4-y-1+-sqrt25
x-y=3+-sqrt25

I've never done this before, I'm not sure if this is even right and if it is, I definitely don't know where to go from here

• pre-calculus - ,

no,
the standard equation of a circle with centre (h,k) and radius r is
(x-h)^2 + (y-k)^2 = r^2

so we ended up with
(x+4)^2 + (y-1)^2 = 25 or
(x+4)^2 + (y-1)^2 = 5^2

then the centre is (-4,1) and radius is 5

Here is an interesting page for your topic
http://www.analyzemath.com/CircleEq/Tutorials.html

• pre-calculus - ,

Thanks so much for your explanation. I was able to figure it out and now got your answer.