Sure, let's convert the given equation of the circle from its general form to its standard form step-by-step.
Given equation:
[tex]\[ x^2 + y^2 + 8x + 22y + 37 = 0 \][/tex]
To convert this to the standard form of a circle, [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we need to complete the square for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
1. Complete the square for the [tex]\(x\)[/tex]-terms:
- Start with [tex]\( x^2 + 8x \)[/tex].
- Take half of 8, which is 4, and square it to get 16.
- Rewrite [tex]\( x^2 + 8x \)[/tex] as [tex]\( (x + 4)^2 - 16 \)[/tex].
2. Complete the square for the [tex]\(y\)[/tex]-terms:
- Start with [tex]\( y^2 + 22y \)[/tex].
- Take half of 22, which is 11, and square it to get 121.
- Rewrite [tex]\( y^2 + 22y \)[/tex] as [tex]\( (y + 11)^2 - 121 \)[/tex].
3. Substitute these completed squares back into the original equation:
[tex]\[ (x + 4)^2 - 16 + (y + 11)^2 - 121 + 37 = 0 \][/tex]
4. Combine the constants:
[tex]\[ (x + 4)^2 + (y + 11)^2 - 100 = 0 \][/tex]
[tex]\[ (x + 4)^2 + (y + 11)^2 = 100 \][/tex]
So, the equation of the circle in standard form is:
[tex]\[ (x + \boxed{4})^2 + (y + \boxed{11})^2 = \boxed{100} \][/tex]
5. Identify the center and radius of the circle:
- The center of the circle is at the point [tex]\((h, k)\)[/tex] which is [tex]\((-4, -11)\)[/tex].
- So, the center of the circle is at the point [tex]\( (\boxed{-4} , \boxed{-11}) \)[/tex].
Therefore, the completed answer is:
[tex]\[ \text{The equation of this circle in standard form is } (x + \boxed{4})^2 + (y + \boxed{11})^2 = \boxed{100} \][/tex]
[tex]\[ \text{The center of the circle is at the point } (\boxed{-4}, \boxed{-11}) \][/tex]
