To convert the equation of the circle [tex]\( x^2 + y^2 + 8x + 22y + 37 = 0 \)[/tex] into standard form, follow these steps:
1. Complete the square for the [tex]\(x\)[/tex] terms and [tex]\(y\)[/tex] terms separately:
For the [tex]\(x\)[/tex] terms: [tex]\( x^2 + 8x \)[/tex]
- Take the coefficient of [tex]\(x\)[/tex], which is 8, and divide it by 2, getting 4. Then, square this result to get 16.
- Write the [tex]\(x\)[/tex] terms as [tex]\( (x + 4)^2 - 16 \)[/tex].
For the [tex]\(y\)[/tex] terms: [tex]\( y^2 + 22y \)[/tex]
- Take the coefficient of [tex]\(y\)[/tex], which is 22, and divide it by 2, getting 11. Then, square this result to get 121.
- Write the [tex]\(y\)[/tex] terms as [tex]\( (y + 11)^2 - 121 \)[/tex].
2. Rewrite the original equation with these completed squares:
[tex]\[
(x + 4)^2 - 16 + (y + 11)^2 - 121 + 37 = 0
\][/tex]
3. Combine and simplify the constant terms on the left-hand side:
[tex]\[
(x + 4)^2 + (y + 11)^2 - 16 - 121 + 37 = 0
\][/tex]
[tex]\[
(x + 4)^2 + (y + 11)^2 - 100 = 0
\][/tex]
4. Move the constant term to the right-hand side:
[tex]\[
(x + 4)^2 + (y + 11)^2 = 237
\][/tex]
Thus, the equation in standard form is:
[tex]\[
(x + \boxed{4})^2 + (y + \boxed{11})^2 = \boxed{237}
\][/tex]
The center of the circle is at the point:
[tex]\(
(\boxed{-4}, \boxed{-11})
\)[/tex]
