Answer:
C. Center (3,-2) and radius 3
Step-by-step explanation:
The given circle has equation:
[tex]x^2+y^2-6x+4y+4=0[/tex]
We rearrange to get:
[tex]x^2-6x+y^2+4y=-4[/tex]
We add the square of half the coefficients of the linear terms to both sides of the equation:
[tex]x^2-6x+(-3)^2+y^2+4y+(2)^2=-4+(-3)^2+(2)^2[/tex]
We factor the perfect squares and simplify to get;
[tex](x-3)^2+(y+2)^2=9[/tex]
We can rewrite as;
[tex](x-3)^2+(y--2)^2=3^2[/tex]
Comparing this to the standard equation of the circle;
[tex](x-h)^2+(y-k)^2==r^2[/tex]
We have (3,-2) and the center and r=3 as the radius.
