What is the radius for the circle given by the equation x^2+(y-1)^2=12?

Question

tkhunny · Answer

Eyeball problem.  Just look at it and write down the answer.  $\sqrt{12}$

jhannybean · Answer

$$(x-h)^2 +(y-k)^2 = r^2$$$$\text{center} = (h~,~k)$$$$\text{radius}=\sqrt{r^2}=r$$

ria23 · Answer

In this case, would my h be 1?

zale101 · Answer

(h,k) we don't know the numbers that indicates the center it just gives the variables, the only number we have is the radius. If it was
$$ (x-1)^2+(y-k)^2= \sqrt{12}$$
then h would be 1

tkhunny · Answer

Why do you care about the center of the circle.  It asks only for the radius.

ria23 · Answer

First, I wanna point out that I'm not out for the answer
and second, I'm sorry but I am so confused. >~<

would my equation end up being $$(x-1)^{2}+(x+1)^{2}=\sqrt12$$?

jhannybean · Answer

Its not a matter of caring, it's simply defining what the formula gives, and how it relates to the question given.

jhannybean · Answer

@Ria23 your equation does not change, but the question you posted was simply asking for the radius of the circle. Idk why you changed the format of your problem.

jhannybean · Answer

I was simply relating the format of the equation of the circle and your equation :$$\color{blue}{(x-h)^2} +\color{red}{(y-k)^2} = \color{violet}r^2$$$$\color{blue}{(x+0)^2}+\color{red}{(y-1)^2}=\color{violet}{12}$$

jhannybean · Answer

So when you equate $r^2 = 12$ you get $r= \pm\sqrt{12}$ but since the radius is always positive, you will have $r = \sqrt{12}$

ria23 · Answer

Oh... Oh, I get it now. 
Thank yhu for explaining that for me. I wasn't sure what everyone meant...

jhannybean · Answer

No problem. The center is just centered at your point (h, k) which in your case would be (0,1). That was just extra information.

ria23 · Answer

Can yhu explain why that would be 0?
I thought that if yhu have a variable without a number in front it automatically became a 1...

jhannybean · Answer

It's not the number in front that matters. It is 0 because your original equation states : $x^2+(y-1)^2=12$ 

So in order to fit THIS equation into the format of the equation of a circle, I set it up like my colorful post above. Therefore I found that the x-component for the center is 0.

jhannybean · Answer

Does that make sense?

ria23 · Answer

Yes it does. c: Thank yhu again for all yhur help! ^.^

jhannybean · Answer

Woo! awesome :)