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The general form of the equation of a circle is [tex]x^2 + y^2 + 8x + 22y + 37 = 0[/tex].

The equation of this circle in standard form is [tex](x + \square)^2 + (y + \square)^2 = \square[/tex].

The center of the circle is at the point ([tex]\square[/tex], [tex]\square[/tex]).


Sagot :

Let's start with the given equation of the circle in general form:

[tex]\[ x^2 + y^2 + 8x + 22y + 37 = 0 \][/tex]

To convert this to the 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]:

[tex]\[ x^2 + 8x \][/tex]

To complete the square, add and subtract [tex]\( 16 \)[/tex] (since [tex]\((\frac{8}{2})^2 = 16\)[/tex]):

[tex]\[ x^2 + 8x = (x + 4)^2 - 16 \][/tex]

2. Complete the square for [tex]\( y \)[/tex]:

[tex]\[ y^2 + 22y \][/tex]

To complete the square, add and subtract [tex]\( 121 \)[/tex] (since [tex]\((\frac{22}{2})^2 = 121\)[/tex]):

[tex]\[ y^2 + 22y = (y + 11)^2 - 121 \][/tex]

3. Rewrite the original equation using the completed squares:

[tex]\[ x^2 + y^2 + 8x + 22y + 37 = 0 \][/tex]

Substitute the completed squares:

[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]

5. Isolate the squared terms:

[tex]\[ (x + 4)^2 + (y + 11)^2 = 100 \][/tex]

Therefore, the equation 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].

The radius of the circle is:

[tex]\[ \sqrt{100} = 10 \][/tex]

So, filling in the boxes, we get:

The equation of this circle in standard form is:

[tex]\[ (x + \square)^2 + (y + \square)^2 = \square \][/tex]

with the boxes filled in as follows:

1. [tex]\(\square\)[/tex]: -4
2. [tex]\(\square\)[/tex]: -11
3. [tex]\(\square\)[/tex]: 100

The center of the circle is at the point:

[tex]\[ (\square, \square) \][/tex]

with the boxes filled in as follows:

1. [tex]\(\square\)[/tex]: -4
2. [tex]\(\square\)[/tex]: -11