Answer
$(x-5)^2+(y-3)^2=64$
$C(5,3), r=8$
See graph.
Work Step by Step
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.
