ANSWER
[tex]\begin{gathered} center=(-3,-5) \\ r=4 \end{gathered}[/tex]
EXPLANATION
Given;
[tex]x^2+y^2+6x+10y+18=0[/tex]
The standard equation of circle of a circle centered at (h,k) with radius r is;
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Re-write the given equation in the standard form, we have;
[tex]\begin{gathered} x^{2}+y^{2}+6x+10y+18=0 \\ x^{2}-6x+(\frac{6}{2})^{2}+y^{2}+10y+(\frac{10}{2})^{2}=-18+9+25 \\ (x-3)^2+(y-5)^2=4^2 \end{gathered}[/tex]
Hence, h=-3, k=-5, radius is 4
