Question

Can somebody help me with this question, it has to do with graphing circles.

Answer 1

go to walmart and buy yourself a compass

Answer 2

Haha, yhur so funny. -_-

Answer 3

\[(x-h)^2 + (y-k)^2 = r^2\]

Answer 4

(h,k) is center
r = radius

Answer 5

So, how would I set up the equation?

Answer 6

in the equation you can see \(x^2\) , that means h = 0 in the equation given by dumbcow.
you also see \((y+3)^2=(y-(-3))^2\), that means \(k=-3\).
-> center is \((0,-3)\).
that's what the LHS tells you.

the RHS gives the radius.

Answer 7

Ok, so I know my answer... But I'm not quite sure... How I find the radius...
So I have the center... I don't... oy. circles. I don't know how I find the radius... my teacher either squared it, or took the square root to get it... Se just kind of pulled numbers... I don't know how to find it...

Answer 8

on the left side, everything is "already as as it should" = "in the form \((x-h)^2+(y-k)^2\)" thanks to this you can tell what the center of the circle is.

When "things are right" on the left hand side, the right hand side IS equal to \(r^2\). Here, you read 9. From there you know that the radius is \(\sqrt9=3\).

actually, in the options A,B,C,D, there's only one of the circles who has center (0,-3) so you didn't really need to find the radius.

Answer 9

I know c: I just didn't know how I would find the radius in these kinds of equations. So, I do take the square root? Or is that only for certain cases?

Answer 10

if you're dealing with circle equations, the equation will always "display" \(r^2\).

Answer 11

Oh. Alright.
Thank yhu so much for yhur help reemii and @dumbcow