## idreamofpink Group Title NEED HELP WITH PRE CAL QUESTION (PIC INSERTED) one year ago one year ago

1. idreamofpink

2. tkhunny

Did you learn "Completing the Square"? You will need that.

3. RaphaelFilgueiras

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4. RaphaelFilgueiras

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5. tkhunny

Really, @RaphaelFilgueiras ? You're just going to drop an answer with absolutely no explanation? Please rethink this behavior.

6. RaphaelFilgueiras

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7. 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.

8. 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

9. theEric

and factor wasn't the right term, I think tkhunny got it right with "completing the square!"

10. RaphaelFilgueiras

Now imagine we are in new coordinates x' and y'|dw:1372825398862:dw|

11. 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 "$$-3^2$$".

12. 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!!}$

13. 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. $\huge\text{ To the finish!}$

14. 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.

15. theEric

The radius of the circle is $$2$$.

16. theEric

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17. theEric

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18. 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!