<s>[INST] 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 $x^2 + y^2 -6x + 10y + 9 = 0$
 [/INST] A mechanical way is to start with the usual equation of a circle centered at the point $(a,b)$ and radius $r$ 
$$
(x-a)^{2}+(y-b)^{2}=r^{2},\tag{1}
$$
write it in the form
$$
(x-a)^{2}+(y-b)^{2}-r^{2}=0,
$$
and expand the LHS
$$
\begin{eqnarray*}
(x-a)^{2}+(y-b)^{2}-r^{2} &=&\left( x^{2}-2ax+a^{2}\right) +\left(
y^{2}-2by+b^{2}\right) -r^{2} \\
&=&x^{2}+y^{2}-2ax-2by+a^{2}+b^{2}-r^{2},
\end{eqnarray*}
$$
so that we get the equivalent equation 
$$
x^{2}+y^{2}-2ax-2by+a^{2}+b^{2}-r^{2}=0.\tag{2}
$$
Equating the coefficients of $(2)$ to the ones of the given equation
$$
x^{2}+y^{2}-6x+10y+9=0\tag{3}
$$
results in the system of equations 
$$
\begin{eqnarray*}
-2a &=&-6 \\
-2b &=&10 \\
a^{2}+b^{2}-r^{2} &=&9,
\end{eqnarray*}
$$
which is equivalent to
$$
\begin{eqnarray*}
a &=&3 \\
b &=&-5 \\
9+25-r^{2} &=&9.
\end{eqnarray*}
$$
So the center is the point $(a,b)=(3,-5)$ and the radius is $r=5$, because it cannot be negative (the other solution of $9+25-r^{2} =9$ is $-5$).
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