anonymous
  • anonymous
NEED HELP WITH PRE CAL QUESTION (PIC INSERTED)
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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tkhunny
  • tkhunny
Did you learn "Completing the Square"? You will need that.
anonymous
  • anonymous
|dw:1372824594206:dw|

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anonymous
  • anonymous
|dw:1372824729631:dw|
tkhunny
  • tkhunny
Really, @RaphaelFilgueiras ? You're just going to drop an answer with absolutely no explanation? Please rethink this behavior.
anonymous
  • anonymous
|dw:1372825327340:dw|
theEric
  • theEric
Hi! Sorry I didn't get to this while you were around. I had to refresh my memory, and organize the technique so I could teach it to you. You might know it! In case you don't, I'll write it out. You have to do it twice, and then the problem is \(almost\) solved. \[\huge\mathrm{\text{Finding a Better Equation:}}\] First you have to find an equation that is easier to pick information from. If your a very capable numbers person, you might see that equation and say, "circle!" But most of us aren't like that. Here's a technique that I KNOW will help you here. Look at your \((x^2+6x)\). That's asking to be changed. Hopefully in a way to get you to the circle formula, which is \((x+a)^2+(y+b)^2=r^2\) where \(a\), and \(b\) are any real numbers. For that \((x^2+6x)\), lets turn it into something bigger. You'll have to trust me at first. It should look like a polynomial (\(ax^2 + bx +c\) where \(a\), \(b\), and \(c\) are any real numbers). By doing so, we can turn it into something that we can - and I think I'm using the right term here - factor. \(x^2+6x\) a.k.a. \(x^2+6x+0\) is not something we can just factor. For this type of polynomial (second order polynomial) it'll be in a form \(a^2 + 2ab + b^2\) which equals \((a+b)^2\). Again, \(a\) and \(b\) are real numbers. We have the \(a^2 +2ab\) part, really.
theEric
  • theEric
Ah, circle equation is actually \[(x-a)^2+(y-b)^2=r^2\] And I didn't see other people were answering this, so sorry if everything is jumbled! :P
theEric
  • theEric
and factor wasn't the right term, I think tkhunny got it right with "completing the square!"
anonymous
  • anonymous
Now imagine we are in new coordinates x' and y'|dw:1372825398862:dw|
theEric
  • theEric
Continuing on with my explanation, in case you choose to read it or other answers aren't completed, we were at \(x^2 + 6x\) actually being the first part of our polynomial, the \(a^2 + 2ab\) part. The whole thing was \(a^2 + 2ab + b^2\). \[\Huge\text{Continuing making it}\\\huge\text{look better with}\\\Huge\text{"Completing the Square"}\] So, if \(x^2+6x=a^2+2ab\), you can logic out what \(b^2\) is. Let's assume that \(x=a\). Then \(x^2=a^2\) Well, that part's taken care of. Since \(x^2+6x=a^2+2ab\) and \(x^2 = a^2\), guess what we can do next! Subtract that value from both sides. It's okay. If we treat both sides equally, they'll still be equal. \(\cancel{x^2}-\cancel{x^2}+6x=\cancel{a^2} - \cancel{x^2}+2ab\). So, \(6x=2ab\). Hmmm... \(a=x\), right? Then we can divide by that value. Then we see that \(\Large\frac{6\cancel{x}}{\cancel{x}}=\frac{2\cancel{a}b}{\cancel{x}}\). Cleaned up, to see it clearer, we have \(6=2b\). Divide both sides by \(2\), now, so \(b\) is alone. \(\Large\frac{6}{2}=\frac{\cancel{2}b}{\cancel{2}}\normalsize =3=b\). Just like that. That technique gave us \(x=a\) and \(3=b\). That's for \(a^2+2ab+b^2\). Here's the thing. We only had \(x^2+6x\) to start with, and the polynomial we want is \(x^2+6x+3^2\). We can't add \(3^2\) out of nowhere... Unless we subtract it at the same time... So, we got from \(x^2 +6x\) to \(x^2+6x+3^2 - 3^2\). And now, we can complete the square easily. \(a^2 + 2ab + b^2 = (a+b)^2\). Likewise,\[x^2+6x+3^2 -3^2=(x+3)^2-3^2\]Notice how we have to maintain the "\(-3^2\)".
theEric
  • theEric
We just completed the square, I guess. But we need to do the same for \(y-8y\). \[\Huge\text{"Completing the Square"} \\ \Large\text{again}\] *** I \(\it{highly}\) recommend you either try this for yourself first, try it afterwards, or both. There's no way to know if you absorbed it without trying. *** Let \(y\) be \(a\) in \(a^2 + 2ab +b^2\). So \(y=a\). Then \(y^2=a^2\). To \(y^2 -8y = a^2+2ab\), we subtract the \(y^2=a^2\) value. We get \(-8y=2ab\). We divide that by the \(y=a\) value. We get \(-8=2b\). We divide by 2. We get \(\Large\frac{-8}{2}\normalsize =-4=b\). To \(y^2-8y\), we add and subtract \(b^2 = 16\). We get \(y^2-8y+16-16\). We use the formula \(a^2+2ab+b^2=(a+b)^2\) on the previous expression. We get \(y^2-8y+16-16=\left(y+\left(-4\right)\right)^2-16=(y-4)^2-16\). \[\huge\text{We insert back into} \]\[ \huge \text{the original equation!!}\]
theEric
  • theEric
We had\[x^2+y^2+6x-8y+21=0\]\[x^2+6x+y^2-8y+21=0\] We added and subtracted \((3)^2=9\) and \((-4)^2=16\), getting\[\begin{matrix} x^2+6x+9& +& y^2-8y+16 &+& 21-9-16&=&0\end{matrix}\]This simplifies to \begin{matrix} (x+3)^2&&+&&(y-4)^2&&+&&(-4)&&=&&0\end{matrix} Now the new - add 4 to both sides to get\[(x+3)^2+(y-4)^2=4=2^2\]\[\large\text{And so we have the circle formula,}\]\[(x-a)^2+(y-b)^2=r^2\]Again, \(a\) and \(b\) are just some real numbers. \[\huge\text{ To the finish!}\]
theEric
  • theEric
The easiest way to understand the circle formula is memorization. In \((x-a)^2+(y-b)^2=r^2\), \(r\) is your radius \(a\) is how much the circle's center shifts \(\boldsymbol{right}\) from the origin \(b\) is how much the circle's center shifts \(\boldsymbol{up}\) from the origin. Our more understandable formula is \((x+3)^2+(y-4)^2=2^2\). That's like, \((x-(-3))^2+(y-4)^2=2^2\) Our \(a\) is \(-3\) and our \(b\) is \(4\). Since \(a\) is negative, the center of of the circle shifts the opposite of right: left.
theEric
  • theEric
The radius of the circle is \(2\).
theEric
  • theEric
|dw:1372830505247:dw|
theEric
  • theEric
|dw:1372830814043:dw|
theEric
  • theEric
I hope this helps. It might be long, and I'm sorry for that. I try to leave nothing out. I remember trying to solve math problems based on examples, and I'd sometimes not be able to see how people got from step 1 to step 2. Good luck!

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