SAT Question b96ff36e: Circles answer and explanation with Desmos solution

It's given that in the xy-plane, the graph of the equation $(x-3{)}^2+(y-5{)}^2=9$ is a circle. It's also given that the point $(6,c)$, where $c$ is a constant, lies on this circle. It follows that the ordered pair $(6,c)$ makes the equation $(x-3{)}^2+(y-5{)}^2=9$ true. Substituting $6$ for $x$ and $c$ for $y$ in this equation yields $(6-3{)}^2+(c-5{)}^2=9$, or $9+(c-5{)}^2=9$. Subtracting $9$ from each side of this equation yields $(c-5{)}^2=0$. It follows that the value of $c$ is $5$.