\(The\ equation\ of\ a\ circle:(x-a)^2+(y-b)^2=r^2\\\\(a;\ b)-a\ coordinates\ of\ the\ center\\r-a\ radius\)
\(x^2-6x+y^2+10y+18=0\\\\x^2-2x\cdot3+y^2+2y\cdot5=-18\\\\x^2-2x\cdot3+3^2+y^2+2y\cdot5+5^2=-18+3^2+5^2\ \ \ |use\ (a\pm b)^2=a^2\pm2ab+b^2\\\\(x-3)^2+(y+5)^2=-18+9+25\\\\(x-3)^2+(y+5)^2=16\\\\therefore\\\\Answer:\boxed{r=\sqrt{16}=4}\)
