Question

show that x^2+ax+y^2+by+c=0 is the equation of a circle if and only if a^2+b^2-4c>0

Answer 1

use completing the square to put equation of circle in standard form
(x+a/2)^2 + (y+b/2)^2 = (a^2/4)+(b^2/4)-c
r^2 = (a^2/4)+(b^2/4)-c > 0
multiply equation by 4
-> a^2 +b^2 -4c > 0

Answer 2

ok so am curious, why must it be greater than 0 ?

Answer 3

the square of a number is always positive thus greater than 0.
also a circle must have a positive radius

Answer 4

ok cool ty for the help