Question

If the graph of the given equation is a circle, find its center and radius. If the equation has no graph, say so.

Answer 1

\[x^2+y^2=-8y\]

Answer 2

@terenzreignz

Answer 3

Ahh... the joys of analytic geometry.
More like woes, in your case, but that's why I'm here, to turn them into joys ^_^

Okay, every equation of a circle can be written as follows, yes?
\[\Large (x-\color{red}h)^2 + (y-\color{green}k)^2 = \color{blue}r^2\]

Where \(\large (\color{red}h,\color{green}k) \) is the centre, and \(\large \color{blue} r\) is the radius.

Answer 4

yes so I had so far.... \[x^2 + (y^2+8y+?)=0\]

Answer 5

Very good, <what was your name again?>.
Now, I need you to possess a certain skill... it's called "completing the square"
Have you heard of it? ^_^

Answer 6

hello! hahaha, yes but I kind of have trouble with it, or need to refresh my memory...

Answer 7

Completing the Square 101
Imagine you have something of the form
\[\Large z^2 + \color{purple}pz = k\]
And for some reason or another, you need to express it in terms of a perfect square trinomial.

The first thing to watch out for is to make sure that the square bit... (this part)\[\Large \color{red}{z^2 }+ \color{purple}pz = k\] is unadorned... meaning, it has

no number next to it... if it does, well, remedy it by dividing the entire expression by that number...

So now, pay attention to the coefficient of the 'unsquared' variable, in this case, \(\large \color{purple}p\). What you need to do is halve it, resulting in \(\Large

\frac{\color{purple}p}{2}\) and then square it, yielding \(\Large \frac{\color{purple}p^2}{4}\).

Next, you add this to both sides of your equation, like so..
\[\Large z^2 + \color{purple}pz+\frac{\color{purple}p^2}4 = k+\frac{\color{purple}p^2}{4}\]

You'll notice that the left side is now a perfect square, in particular, the square of \(\Large z+\frac{\color{purple}p}2\)...
\[\Large \left(z+\frac{\color{purple}p}2\right)^2=k+\frac{\color{purple}p^2}{4}\]

And that's it!
You have successfully 'completed the square'
:)

Answer 8

Read and understand, and then I want you to complete the square of the part
\[\Large y^2 + 8y\]

Answer 9

And if you DON'T understand, tell me ASAP :)

Answer 10

okay hold on for a bit.

Answer 11

so why do I halve P before I square it, why not just square it?

Answer 12

Good question.
It's because a perfect square trinomial, say, we square a + qb
it'd look like
\[\Large (a+qb)^2 = a^2 + \color{orange}{2q}ab + \color{green}{q^2}b^2\]

And notice that the orange coefficient is just twice the square root of green coefficient

or equivalently

The green coefficient is the square of HALF the orange coefficient.

Clear? ^_^

Answer 13

okay haha it's a little bit confusing but I think I get it.

Answer 14

Now, get back to work >:)

Answer 15

Complete the square in \[\Large y^2 + 8y\]
in case you get lost :P

Answer 16

okay one sec

Answer 17

so then i get \[y^2 + 8y+16\]

Answer 18

Great ^_^
So do we just do this?
\[\Large x^2 + y^2 + 8y \color{blue}{+16}=0\]?

Answer 19

yes.. and we group so, \[x^2 + (y^2+8y+16)=0 \]

Answer 20

Nope. :P

Answer 21

Stop right there and LISTEN. (first) ^_^

Answer 22

We can't just do this.
\[\Large x^2 + y^2 + 8y \color{red}{+ 16 }= 0\]

It's not 'fair' to the equation.
We just added 16 to the left-side, we have to do it to the right-side too :P
\[\Large x^2 + y^2 + 8y \color{red}{+16}= 0 \color{blue}{+16}\]

Understood?
^_^

Answer 23

yes hahaha That was what I was gonna do next! :D

Answer 24

I was supposed to do that first then group... lol oops

Answer 25

Could have... would have... didn't :P

Anyway, proceed as thus.

Answer 26

\[x^2+(y+4)^2=16\]

Answer 27

Impressive.
You learn quickly ^_^
Now compare with
\[\Large (x-\color{red}h)^2 + (y-\color{green}k)^2 = \color{blue}r^2\]

And determine your values for \(\color{red}h\), \(\color{green}k\), and \(\color{blue}r\).

Answer 28

which means C=(0,-4) and R= 4 ?

Answer 29

h is 0, k is -4 r=4

Answer 30

What else could I say?

Well done ^_^

Answer 31

Awesome! Thanks again! might need more help in the near future with new material lol... if you're available... I'll tag you! :D

Answer 32

Bloody spoiled ..... >:(



...lol

JK.

Better hope I'm still online by then. I *am* getting sleepy :P