To convert the general form of the circle equation [tex]\( x^2 + y^2 + 8x + 22y + 37 = 0 \)[/tex] into standard form, we need to complete the square for both [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
1. Complete the square for [tex]\( x \)[/tex]:
Start with the [tex]\( x \)[/tex]-terms: [tex]\( x^2 + 8x \)[/tex].
To complete the square:
[tex]\[
x^2 + 8x = (x + 4)^2 - 16
\][/tex]
2. Complete the square for [tex]\( y \)[/tex]:
Next, tackle the [tex]\( y \)[/tex]-terms: [tex]\( y^2 + 22y \)[/tex].
To complete the square:
[tex]\[
y^2 + 22y = (y + 11)^2 - 121
\][/tex]
3. Rewrite the original equation including the completed squares:
Substitute the completed squares back into the equation:
[tex]\[
x^2 + y^2 + 8x + 22y + 37 = (x + 4)^2 - 16 + (y + 11)^2 - 121 + 37 = 0
\][/tex]
Combining the constants:
[tex]\[
(x + 4)^2 + (y + 11)^2 - 100 = 0
\][/tex]
4. Isolate the completed squares:
Move the constant term to the other side of the equation:
[tex]\[
(x + 4)^2 + (y + 11)^2 = 100
\][/tex]
So, the standard form of the equation of the circle is [tex]\( (x + 4)^2 + (y + 11)^2 = 100 \)[/tex].
The center [tex]\((h, k)\)[/tex] of the circle can be identified from the standard form [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex]. In this case, [tex]\( h = -4 \)[/tex] and [tex]\( k = -11 \)[/tex].
Finally, the radius squared [tex]\( r^2 \)[/tex] is equal to 100, so the radius [tex]\( r \)[/tex] is [tex]\( \sqrt{100} = 10 \)[/tex].
Therefore, filling in the boxes:
The equation of this circle in standard form is [tex]\( (x + 4)^2 + (y + 11)^2 = 100 \)[/tex].
The center of the circle is at the point [tex]\( (-4, -11) \)[/tex].
