An equation in the form:
[tex](x-a)^2+(y-b)^2=r^2[/tex]is the standard form for the equation of a circle with center (a,b) and radius r. Here we have:
[tex]x^2+y^2-6x+4y+9=0[/tex]Then, group the x and y terms separately and "move" the constant to the right side of the equation:
[tex]x^2-6x+y^2+4y=-9[/tex]Complete the square:
[tex]x^2-6x+9+y^2+4y+4=-9+9+4[/tex]Factor:
[tex](x-3)^2+(y+2)^2=4[/tex]Express the right side as a square:
[tex](x-3)^2+(y-(-2))^2=2^2[/tex]Therefore:
The center is: (3, - 2), the radius is 2
Answer:
[tex]\begin{gathered} \text{Center: (3,-2)} \\ \text{Radius: 2} \end{gathered}[/tex]