## idreamofpink 2 years ago NEED HELP WITH PRE CAL QUESTION (PIC INSERTED)

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!