First we need to convert the given equation to standard form, only then we can find the center and radius of the circle.
[tex] x^{2} + y^{2} +18x+14y+105=0 \\ \\
x^{2} +18x+ y^{2}+14y=-105 \\ \\
x^{2} +2(x)(9)+ y^{2}+2(y)(7)=-105 \\ \\
x^{2} +2(x)(9)+ 9^{2} + [y^{2}+2(y)(7)+7^{2}] =-105+9^{2}+7^{2} \\ \\
(x+9)^{2}+ (y+7)^{2}=25
[/tex]
The standard equation of circle is:
[tex](x-a)^{2} + (y-b)^{2}= r^{2} [/tex]
with center (a,b) and radius = r
Comparing our equation to above equation, we can write
Center of circle is (-9, -7) and radius of the given circle is 5
