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February 1, 2015

February 1, 2015

Posted by **muffy** on Friday, September 18, 2009 at 9:49pm.

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 -
**Reiny**, Friday, September 18, 2009 at 9:54pmbut 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 -
**muffy - still stuck**, Friday, September 18, 2009 at 10:11pmOk. 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 -
**Reiny**, Friday, September 18, 2009 at 10:17pmno,

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 -
**MUFFY**, Saturday, September 19, 2009 at 10:45amThanks so much for your explanation. I was able to figure it out and now got your answer.

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