Answer
$a)$ The equation does not represent a circle and does not have a graph
$b)$ There is no need to obtain the center and radius, because the equation obtained does not represent a circle.
Work Step by Step
$x^{2}+y^{2}+72=12x$
$a)$
Take $12x$ to the left and $72$ to the right:
$x^{2}+y^{2}-12x=-72$
Group the terms with $x$ together:
$(x^{2}-12x)+y^{2}=-72$
Complete the square for the group. To do so, add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=-12$
$\Big[x^{2}-12x+\Big(\dfrac{-12}{2}\Big)^{2}\Big]+y^{2}=-72+\Big(\dfrac{-12}{2}\Big)^{2}$
$(x^{2}-12x+36)+y^{2}=-72+36$
$(x^{2}-12x+36)+y^{2}=-36$
Now, the expression inside the parentheses is a perfect square trinomial. Factor it:
$(x-6)^{2}+y^{2}=-36$
Since no pair of values of $x$ and $y$ satisfies this equation, it doesn't represent a circle and doesn't have a graph.
$b)$
There is no need to obtain the center and radius, because the equation obtained does not represent a circle.
