anonymous
  • anonymous
Find the standard form of the following equation: x^2+y^2+4x+10y+28=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Yes, I need to write the standard form and then graph the circle.
anonymous
  • anonymous
oops i meant \[(x-h)^2+(y-k)^2=r^2\]
anonymous
  • anonymous
first rewrite as \[x^2+4x+y^2+10y=-28\] then complete the square twice do you know how to do that?

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anonymous
  • anonymous
No i don't, I seem to be stuck and not understanding it
anonymous
  • anonymous
ok i can walk you through it, it is not too hard
anonymous
  • anonymous
just the mechanics. not the reasoning lets look at \(x^2+4x\) what is half of \(4\) ?
anonymous
  • anonymous
2
anonymous
  • anonymous
ok so we write \((x+2)^2\) now what is \(2^2\) ?
anonymous
  • anonymous
4
anonymous
  • anonymous
ok good so when we replace the \(x^2+4x\) on the left with \((x+2)^2\) we have to add \(4\) to the other side
anonymous
  • anonymous
\[(x+2)^2+y^2+10y=-28+4\] or \[(x+2)^2+y^2+10y=-24\] now we repeat the process with \(y^2+10y\)
anonymous
  • anonymous
what is half of \(10\) ?
anonymous
  • anonymous
5
anonymous
  • anonymous
so we are going to replace \(y^2+10y\) by \((y+5)^2\) now that is \(5^2\) ?
anonymous
  • anonymous
25
anonymous
  • anonymous
then when we replace \(y^2+10y\) by \((y+5)^2\) we have to add \(25\) to the other side just like before now we have \[(x+2)^2+(y+5)^2=-24+25\] or \[(x+2)^2+(y+5)^2=1\] as we want
anonymous
  • anonymous
Wow you made it seem so simple, thank you! How do I graph it now?
anonymous
  • anonymous
it is a circle with center at \((-2,-5)\) and radius \(1\) draw that
anonymous
  • anonymous
Thank you!
anonymous
  • anonymous
yw

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