we know that
the equation of the circle in standard form is equal to
[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]
where
[tex](h,k)[/tex] is the center of the circle
r is the radius of the circle
we have
[tex]x^{2}+y^{2} +8x+22y+37=0[/tex]
Convert to standard form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex](x^{2}+8x)+(y^{2} +22y)=-37[/tex]
Complete the square twice. Remember to balance the equation by adding the same constants to each side
[tex](x^{2}+8x+4^{2})+(y^{2} +22y+11^{2})=-37+4^{2}+11^{2}[/tex]
[tex](x^{2}+8x+16)+(y^{2} +22y+121)=-37+16+121[/tex]
[tex](x^{2}+8x+16)+(y^{2} +22y+121)=100[/tex]
Rewrite as perfect squares
[tex](x+4)^{2}+(y+11)^{2}=10^{2}[/tex]
the center is the point [tex](-4,-11)[/tex]
the radius is [tex]r=10\ units[/tex]
therefore
The answer is
a) the equation in standard form is [tex](x+4)^{2}+(y+11)^{2}=10^{2}[/tex]
b) the center of the circle is the point [tex](-4,-11)[/tex] and the radius of the circle is [tex]r=10\ units[/tex]
see the attached figure to better understand the problem
