Question

The general form of the equation of a circle is x2 + y2 + 8x + 22y + 37 = 0.

Answer 1

we know that

the equation of the circle in standard form is equal to

$(x-h)^{2}+(y-k)^{2}=r^{2}$

where

$(h,k)$ is the center of the circle

r is the radius of the circle

we have

$x^{2}+y^{2} +8x+22y+37=0$

Convert to standard form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}+8x)+(y^{2} +22y)=-37$

Complete the square twice. Remember to balance the equation by adding the same constants to each side

$(x^{2}+8x+4^{2})+(y^{2} +22y+11^{2})=-37+4^{2}+11^{2}$

$(x^{2}+8x+16)+(y^{2} +22y+121)=-37+16+121$

$(x^{2}+8x+16)+(y^{2} +22y+121)=100$

Rewrite as perfect squares

$(x+4)^{2}+(y+11)^{2}=10^{2}$

the  center is the point $(-4,-11)$

the radius is $r=10\ units$

therefore

The answer is

a) the equation in standard form is $(x+4)^{2}+(y+11)^{2}=10^{2}$

b) the  center of the circle is the point $(-4,-11)$ and the radius of the circle is $r=10\ units$

see the attached figure to better understand the problem

Answer 2

Hello,

x²+y²+8x+22y+37=0
==>(x²+2*4x+16)+(y²+2*11y+121)+37-16-121=0
==>(x+4)²+(y+1)²=10²
Center=(-4,-1) , radius=10