Question

What is the equation of this circle?
Answer

x2 + (y – 1)2 = 4

x2 + (y + 1)2 = 4

x2 + (y – 1)2 = 2

x2 + (y + 1)2 = 2

Answer 1

What are the given pieces of information?

Answer 2

A circle is shown on the coordinate plane below.

Circle that contains points 2, 1; 0, 3; negative 2, 1; 0, negative 1.

Answer 3

Because is is just x^2, the x coordinate of the center of the circle must be 0. Which point(s) have that?

Answer 4

0, 3 and 0, -1

It can't be 0,3 because then the equation of that circle will be of the form x^2 + (y-3)^2 = r^2.

So, it has to be 0, -1.

Answer 5

thank you once again u should be a mod XD

Answer 6

I don't need the moderator power. :)

Answer 7

If (0,3) is on the circle and (0,-1) is on the circle, the center of the circle must be half way between. So the center is (0,1) and the equation of the circle is x^2+(y-1)^2 = 4

Answer 8

I am confused. Were those the choices for centers of circles are "points ON" the circle?

Answer 9

??

Answer 10

idk

Answer 11

I do not see how this statement is true:
If (0,3) is on the circle and (0,-1) is on the circle, the center of the circle must be half way between.

Answer 12

yes i see how its true r u talking to me or Mertsj