anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
divide by 2 first and get \[x^2+2x+y^2-4y=11\]
anonymous
  • anonymous
then complete the square twice \[(x+1)^2+(y-2)^2=11+1+4\] and you are pretty much done
anonymous
  • anonymous
Im confused though.

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anonymous
  • anonymous
how did you get (x+1)^2 from x^2+2x?
anonymous
  • anonymous
Deal with the x terms first: \[x^2+2x=x^2+2x+1-1\\ ~~~~~~~~~~~~=(x+1)^2-1\]
anonymous
  • anonymous
wheres the one coming from
anonymous
  • anonymous
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."
anonymous
  • anonymous
\[x^2+2x=x^2+2x+0\\~~~~~~~~~~~~=x^2+2x+(1-1)\\~~~~~~~~~~~~=x^2+2x+1-1\\~~~~~~~~~~~~=(x+1)^2-1\]
anonymous
  • anonymous
what do you do next then
anonymous
  • anonymous
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.

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