Question

Find the center, radius, and intercepts.
2x^2+2y^2+8x+7=0

Answer 1

I need help on putting this in Standard Form. From there, I believe I can do the rest.

Answer 2

you gotta complete the square twice
once for the x terms and once for the y terms

Answer 3

I did and got 2(x+2)^2+2(y)^2=1
I'm not sure if this is correct though.

Answer 4

\[2x^2+2y^2+8x+7=0\] subtract \(7\)\[2x^2+2y^2+8x=-7\] divide by 2 to make the leading coefficient 1 and start with
\[x^2+4x+y^2=-\frac{7}{2}\]

Answer 5

it is not correct, you can't complete the square without making the leading coefficient one

Answer 6

standard form
\[(x-h)^2+(y-k)^2=r^2\]

Answer 7

Oh! Thank you!

Answer 8

you good from there? how to complete the square i mean?

Answer 9

Yes! I just have one last question, how can you tell if you have x-int and y-int?

Answer 10

x intercept if what you get when you put \(y=0\)

Answer 11

you can do that without putting it in standard form, just get rid of the y and solve for \(x\)

Answer 12

but you need the center and radius anyway, so you have no choice but to complete the square

Answer 13

Oh yeah! :D Thank you!