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Translate the following conic from standard form to graphing form:

[tex]\[ x^2 - 6x + y^2 - 4y + 12 = 0 \][/tex]

A. [tex]\((x-2)^2 + (y+5)^2 = 11\)[/tex]

B. [tex]\((x+1)^2 + (y-4)^2 = 16\)[/tex]

C. [tex]\((x-3)^2 + (y-2)^2 = 1\)[/tex]

D. [tex]\((x+2)^2 + (y-1)^2 = 9\)[/tex]


Sagot :

To convert the given equation from standard form to graphing form, we will use the method of completing the square.

The given equation is:
[tex]\[ x^2 - 6x + y^2 - 4y + 12 = 0 \][/tex]

### Step-by-Step Solution

1. Group the [tex]\(x\)[/tex] terms and the [tex]\(y\)[/tex] terms together:
[tex]\[ (x^2 - 6x) + (y^2 - 4y) + 12 = 0 \][/tex]

2. Complete the square for the [tex]\(x\)[/tex] terms:
[tex]\[ x^2 - 6x = (x - 3)^2 - 9 \][/tex]

3. Complete the square for the [tex]\(y\)[/tex] terms:
[tex]\[ y^2 - 4y = (y - 2)^2 - 4 \][/tex]

4. Substitute these squares back into the equation:
[tex]\[ (x - 3)^2 - 9 + (y - 2)^2 - 4 + 12 = 0 \][/tex]

5. Simplify the constant terms:
[tex]\[ (x - 3)^2 + (y - 2)^2 - 9 - 4 + 12 = 0 \][/tex]
[tex]\[ (x - 3)^2 + (y - 2)^2 - 1 = 0 \][/tex]

6. Move the constant term to the right side of the equation:
[tex]\[ (x - 3)^2 + (y - 2)^2 = 1 \][/tex]

The converted equation is now in graphing form:
[tex]\[ (x - 3)^2 + (y - 2)^2 = 1 \][/tex]

This matches the option:
c [tex]\((x - 3)^2 + (y - 2)^2 = 1\)[/tex]

Thus, the correct answer is:
[tex]\[ \boxed{\text{c}} \][/tex]
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