Chapter 13 - Conic Sections - 13.1 Conic Sections: Parabolas and Circles - 13.1 Exercise Set - Page 854: 35

$(x-7)^2+(y-3)^2=6$

Using $(x-h)^2+(y-k)^2=r^2$ or the Center-Radius form of the equation of circles, the equation of the circle with center $( 7,3 )$ and radius, $r= \sqrt{6} ,$ is \begin{array}{l}\require{cancel} (x-7)^2+(y-3)^2=\left( \sqrt{6} \right)^2 \\\\ (x-7)^2+(y-3)^2=6 .\end{array}