Question

How do I convert the equation of this circle from general form to standard form?
2x^2 +2y^2 +4x -8y -22 =0
(I know I must complete the square, but I am stuck...)

Answer 1

standard form is (x-h)^2 + (y-k)^2= r^2 I know this...
how should I start?

Answer 2

2x^2 +4x +___ +2y^2 -8y + ___ -22 =0 ???

Answer 3

can someone help please?

Answer 4

\(\bf 2x^2 +2y^2 +4x -8y -22 =0\\ \quad \\p
\textit{let us first group, yes}\\ \quad \\
(2x^2 +4x)+(2y^2 -8y) -22 =0\\
\textit{let's take common factors}\\
2(x^2 +2x)+2(y^2 -4y) -22 =0\\ \quad \\
\textit{now let's complete the square}\\
2(x^2 +2x+\square^2)+2(y^2 -4y+\square^2) -22 =0\)

Answer 5

so... what do you think our values would be?

Answer 6

hello, thanks for the help! i'm reading this as we speak, so bear with me!

Answer 7

completing the square means taking the first coefficient of the x^2 and y^2 values, right? so we would take 1 and divide it by 2 and then square it? so the value for each would be 1/4?

Answer 8

darn... typos.. anyhow

Answer 9

so, \(\bf \textit{now let's complete the square}\\
2(x^2 +2x+\square^2)+2(y^2 -4y+\square^2) -22 =0\\ \quad \\
2(x^2 +2(x)(\square)+\square^2)+2(y^2 -2(y)(2)+\square^2) -22 =0\)

so... what do you think are our missing values?

Answer 10

completing the square means... making it like a perfect square trinomial

Answer 11

the middle term in a "perfect square trinomial" is 2 times the non-squared version of the other terms

Answer 12

ok, but I am still confused I think. I'll look at it a bit more and let you know if I've got anything.

Answer 13

\(\huge a^2+2ab+b^2\implies (a+b)^2\)

Answer 14

http://www.youtube.com/watch?v=xGOQYTo9AKY

Answer 15

so... the first would be: 2(x^2 + 2x +4)?
the second: 2( y^2 -4y + 16) ?

Answer 16

That wasn't correct then?
I'll browse this video if you could keep thinking up a way to help explain it to me.