The equation of a circle whose center is at (4, 0) and radius is length 2^(3) is
(X-4)^2 +y^2=2^3
(X+4)^2+y^2=12
(X-4)^2+y^2=12

Question

The equation of a circle whose center is at (4, 0) and radius is length  2^(3) is 
(X-4)^2 +y^2=2^3
(X+4)^2+y^2=12
(X-4)^2+y^2=12

snowflake0531 · Answer

$$(x-h)^2 + (y-k)^2 = r^2$$ where the center is (h,k) and the radius is r

Shay10 · Answer

Okay.

snowflake0531 · Answer

By the way, wdym "the radius is 2^3"

also, are there 4 choices? not 3?

Shay10 · Answer

That's it

snowflake0531 · Answer

Well, based on what I wrote above, which choice do you think it is

Shay10 · Answer

The first one

snowflake0531 · Answer

I think that too

snowflake0531 · Answer

Although it's weird... it should be radius squared... which would make it 2^6, but welp
the first one makes the most sense i suppose

snowflake0531 · Answer

WAIT is it actually $2\sqrt{3}$
???

Shay10 · Answer

I don't understand

Shay10 · Answer

Is this geometry or precalculus

snowflake0531 · Answer

Is it possible for you to give me a screenshot of the question?

snowflake0531 · Answer

@shay10 wrote:

Is this geometry or precalculus

circle stuff is geometry- and you're in the class- how do you not know what class you're taking-

Shay10 · Answer

Idk

snowflake0531 · Answer

well, can you give me a screenshot of the question