Question

This equation of a circle is in the general form. Convert it to standard form (x-h)^2 + (y-h)^2=r^2. The equation is 2x^2+2y^2+4x-8y-22=0

Answer 1

divide by 2 first and get
\[x^2+2x+y^2-4y=11\]

Answer 2

then complete the square twice
\[(x+1)^2+(y-2)^2=11+1+4\] and you are pretty much done

Answer 3

Im confused though.

Answer 4

how did you get (x+1)^2 from x^2+2x?

Answer 5

Deal with the x terms first:
\[x^2+2x=x^2+2x+1-1\\
~~~~~~~~~~~~=(x+1)^2-1\]

Answer 6

wheres the one coming from

Answer 7

I added it to the equation. But I also subtracted. This way, I don't fundamentally change the expression. This is part of "completing the square."

Answer 8

\[x^2+2x=x^2+2x+0\\~~~~~~~~~~~~=x^2+2x+(1-1)\\~~~~~~~~~~~~=x^2+2x+1-1\\~~~~~~~~~~~~=(x+1)^2-1\]

Answer 9

what do you do next then

Answer 10

Similar work with the y terms:
\[\begin{align*}y^2-4y&=y^2-4y+0\\
&=y^2-4y+(4-4)\\
&=y^2-4y+4-4\\
&=(y-2)^2-4\\
\end{align*}\]
So altogether you have
\[(x+1)^2-1+(y-2)^2-4=11\\
(x+1)^2+(y-2)^2=11+1+4\]
just like satellite said earlier.