You wrote correctly

The center is (0,0)

Radius is sqrt72=6sqrt2

But how? That's what I'm lost on.i don't know how to set it up

I got the answer from someone else.i don't know how to set it up to find the center

The standatd formula for the circle with the centre in point (a,b) is

(x-a)^2+(y-b)^2=r^2

We have (x-0)^2+(y-0)^2=72, so the centre is in the point (0,0)

So unless there is a number before X, it's 0?

If there is no number before x and y, then the center has the coordinates (0,0)

Ohhh okay. So how would I go about the numbers? Do I square them all?

no, for example (x-5)^2+(y-2)^2=25 indicate that the center is (5, 2)

to determine the center you just need to see on the values of a and b: (x-a)^2+(y-b)^2=r^2

So if on the test Friday he gives us a problem that says that, we just write those numbers as pair?

yes if the equation will be in such form

But if the equation will be in the form for example (x+5)^2+(y+2)^2=25, the center will be (-5, -2).

So you just need to look at the sign in the (). If it is +, the point is -, and vice versa.

Okay. I think I may have it now

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