Find the center and radius for each of the following:
$x^2 + y^2 - 6x + 10y + 9 = 0$
${
\begin{array}{l}
\textrm{To find center and radius we put the equation in the }\\
\textrm{standard form for a circle by completing the square in}\\
\textrm{both } x \textrm{ and } y:\\\\
\begin{array}{rcl}
x^2+y^2-6x+10y+9 &=& 0\\
(x^2 - 6x + 9) + (y^2 + 10y + 25) &=& 25\\
(x-3)^2 + (y+5)^2 &=& 25
\end{array}\\\\
\textrm{center: } (3,-5)\\
\textrm{radius: }\sqrt{25} = 5
\end{array}}$
$x^2 + y^2 + 6x + 12y - 5 = 0$
${
\begin{array}{l}
\textrm{To find center and radius we put the equation in the }\\
\textrm{standard form for a circle by completing the square in}\\
\textrm{both } x \textrm{ and } y:\\\\
\begin{array}{rcl}
x^2 + y^2 + 6x + 12y - 5 &=& 0\\
(x^2 + 6x + 9) + (y^2 + 12y + 36) &=& 5 + 9 + 36\\
(x+3)^2 + (y+6)^2 &=& 50
\end{array}\\\\
\textrm{center: } (-3,-6)\\
\textrm{radius: }\sqrt{50} = 5\sqrt{2}
\end{array}}$
Find the equation of the circle with center $(-1,5)$ and radius of $6$.
${\displaystyle{(x+1)^2 + (y-5)^2 = 36}}$
Find the equation for the circle with endpoints of a diameter at $(-2,0)$ and $(4,8)$.
${(x-1)^2 + (y-4)^2 = 25}$
Sketch the graph of $x^2+y^2+8x-6y+16=0$, labeling all important aspects.