Question

Precalcus question: find the standard equation, radius and center: x^2+y^2-6x+9=1/36

Answer 1

\[x^2-6x+3^2-3^2+y^2+9=\frac{ 1 }{ 36 }\]
\[\left( x-3 \right)^2-9+y^2+9=\left( \frac{ 1 }{ 6 } \right)^2\]
\[\left( x-3 \right)^2+\left( y-0 \right)^2=\left( \frac{ 1 }{ 6 } \right)^2\]
which is of the form
\[\left( x-h \right)^2+\left( y-k \right)^2=r^2\]
center is (3,0) and radius=1/6