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