Finding the center and radius of a circle given a general degree 2 equation I am trying to find the center and radius of the circle with equation $x^2 + y^2 -6x + 10y + 9 = 0$

You know the equation of a circle has the form
$$
r^2=(x-a)^2+(y-b)^2.
$$
If you have the equation in this form, you can read off the center and radius.
But you don't have that. You need to do a bit of work to get what you have into the form you want.
The tool to use here is:
Completing the Square 
To complete the square for the expression $ax^2+bx+c$ means to find constants $h$ and $k$ such that   $$
     ax^2+bx+c=a(x-h)^2+k.
    $$  To do this, factor out $a$ and add and subtract $b^2\over4a^2$ in the other factor. 

For your equation
 $$\color{maroon}{x^2} +\color{darkgreen}{ y^2} \color{maroon}{-6x }\color{darkgreen}{+ 10y }\color{maroon}{+ 9}=0$$
you want to complete the square on the $\color{maroon}x$ and on the $\color{darkgreen}y$ terms on the left hand side.
For the $y$ terms:
$$
\eqalign{ y^2+10y  &= y^2+10y +{100\over4}-{100\over4} \cr 
                   &=  y^2+10y +25-25\cr
                   &= (y+5)^2-25 
}
$$
The other part of your equation is easier to work out (note its easy when you include the 9 with it):
$$
x^2-6x+9=(x-3)^2
$$
So
$$\eqalign{
& \color{maroon}{x^2} +\color{darkgreen}{ y^2} \color{maroon}{-6x }\color{darkgreen}{+ 10y }\color{maroon}{+ 9}\cr& \iff
\color{maroon}{(x-3)^2}+ \color{darkgreen}{ (y+5)^2-25} =0\cr &\iff (x-3)^2+  (y+5)^2=25.
}
$$
So the center and radius are...
