Question

how do you find the center of a circle whose equation is x^2+y^2-4=0

Answer 1

It's (0,0), because (x-0)^2 = x^2 and (y=0)^2 = y^2.
https://en.wikipedia.org/wiki/Circle#Equations

Answer 2

The equation of a circle with center (h, k) and radius r is:
\( (x - h)^2 + (y - k)^2 = r^2 \)
In your case, h = 0 and k = 0.

Answer 3

Thanks!

Answer 4

wlcm

Answer 5

So in finding the radius what would you do?

Answer 6

Take your given equation. Add 4 to both sides. Then write 4 as the square of 2. Now compare it with the standard equation of a circle I gave you. What is r in your equation?

Answer 7

So would that equation then be set up like this: (x-0)^2+(y-0)^2=2? meaning that=2??

Answer 8

Careful, it is equal 2^2, not just 2.
The radius is 2. You are correct.

Answer 9

Okay Thanks! haha I'm terrible at math to say the least :p