## Precalculus (6th Edition) Blitzer

$(x-5)^2+(y-3)^2=64$ $C(5,3), r=8$ See graph.
Step 1. From the given equation $x^2+y^2-10x-6y-30=0$, we have $(x^2-10x+25)+(y^2-6y+9)=25+9+30$ and $(x-5)^2+(y-3)^2=64$ Step 2. We can identify the center and radius of the circle as $C(5,3), r=8$ Step 3. See graph.