Question

http://i42.tinypic.com/1051y7b.png How do I solve this?

Answer 1

given an equation of circle :
3(x-2)^2 + 3y^2 = 3, it is same with
3(x-2)^2 + 3(y-0)^2 = 3
divide by 3 all sides, giving us
(x-2)^2 + (y-0)^2 = 1

now can u compare with the general equation of circle is
(x-h)^2 + (y-k)^2 = r^2
with (h,k) is the centre and r its radius of circle