The equation of a circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h, k) - a coordinates of the center
r - a radius
[tex] x^2+y^2-6x+4y+4=0\ \ \ |-4\\\\x^2-2\cdot x\cdot3+y^2+2\cdot y\cdot2=-4\ \ \ |+3^2\ |+2^2\\\\\underbrace{x^2-2\cdot x\cdot3+3^2}_{(a-b)^2=a^2-2ab+b^2}+\underbrace{y^2+2\cdot y\cdot2+2^2}_{(a+b)^2=a^2+2ab+b^2}=-4+3^2+2^2\\\\(x-3)^2+(y+2)^2=3^2 [/tex]
Answer:
the center: (3, -2)
the radius: 3
