Answer :
Let's work through completing the square for the given equation of the circle step-by-step to identify the center and the radius correctly.
Given equation:
[tex]\[ x^2 + y^2 + 6x + 4y - 3 = 0 \][/tex]
1. Rearrange terms to isolate [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x^2 + 6x + y^2 + 4y - 3 = 0 \][/tex]
2. Group the [tex]\(x\)[/tex] terms and [tex]\(y\)[/tex] terms together:
[tex]\[ (x^2 + 6x) + (y^2 + 4y) = 3 \][/tex]
3. Complete the square for the [tex]\(x\)[/tex] terms. To do this, take half of the coefficient of [tex]\(x\)[/tex] (which is 6), square it, and add it inside the parenthesis:
[tex]\[ x^2 + 6x \][/tex]
Take [tex]\( \left(\frac{6}{2}\right)^2 = 9 \)[/tex].
So, we rewrite [tex]\(x^2 + 6x\)[/tex] as:
[tex]\[ (x + 3)^2 - 9 \][/tex]
4. Complete the square for the [tex]\(y\)[/tex] terms. Similarly, take half of the coefficient of [tex]\(y\)[/tex] (which is 4), square it, and add it inside the parenthesis:
[tex]\[ y^2 + 4y \][/tex]
Take [tex]\( \left(\frac{4}{2}\right)^2 = 4 \)[/tex].
So, we rewrite [tex]\(y^2 + 4y\)[/tex] as:
[tex]\[ (y + 2)^2 - 4 \][/tex]
5. Substitute these completed square forms back into the original equation:
[tex]\[ (x + 3)^2 - 9 + (y + 2)^2 - 4 = 3 \][/tex]
6. Simplify the equation:
[tex]\[ (x + 3)^2 + (y + 2)^2 - 9 - 4 = 3 \][/tex]
[tex]\[ (x + 3)^2 + (y + 2)^2 - 13 = 3 \][/tex]
[tex]\[ (x + 3)^2 + (y + 2)^2 = 16 \][/tex]
7. Write the equation in standard form:
[tex]\[ (x + 3)^2 + (y + 2)^2 = 4^2 \][/tex]
From this equation, the center of the circle [tex]\((h, k)\)[/tex] is [tex]\((-3, -2)\)[/tex] and the radius [tex]\( r \)[/tex] is 4.
So, the correct completion of the work is:
[tex]\[ (x + 3)^2 + (y + 2)^2 = 4^2 \][/tex]
Therefore, the center is [tex]\((-3, -2)\)[/tex] and the correct completion sentence is:
[tex]\[ \boxed{(x+3)^2+(y+2)^2=4^2,\ \text{so the center is}\ (-3, -2).} \][/tex]
Given equation:
[tex]\[ x^2 + y^2 + 6x + 4y - 3 = 0 \][/tex]
1. Rearrange terms to isolate [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[ x^2 + 6x + y^2 + 4y - 3 = 0 \][/tex]
2. Group the [tex]\(x\)[/tex] terms and [tex]\(y\)[/tex] terms together:
[tex]\[ (x^2 + 6x) + (y^2 + 4y) = 3 \][/tex]
3. Complete the square for the [tex]\(x\)[/tex] terms. To do this, take half of the coefficient of [tex]\(x\)[/tex] (which is 6), square it, and add it inside the parenthesis:
[tex]\[ x^2 + 6x \][/tex]
Take [tex]\( \left(\frac{6}{2}\right)^2 = 9 \)[/tex].
So, we rewrite [tex]\(x^2 + 6x\)[/tex] as:
[tex]\[ (x + 3)^2 - 9 \][/tex]
4. Complete the square for the [tex]\(y\)[/tex] terms. Similarly, take half of the coefficient of [tex]\(y\)[/tex] (which is 4), square it, and add it inside the parenthesis:
[tex]\[ y^2 + 4y \][/tex]
Take [tex]\( \left(\frac{4}{2}\right)^2 = 4 \)[/tex].
So, we rewrite [tex]\(y^2 + 4y\)[/tex] as:
[tex]\[ (y + 2)^2 - 4 \][/tex]
5. Substitute these completed square forms back into the original equation:
[tex]\[ (x + 3)^2 - 9 + (y + 2)^2 - 4 = 3 \][/tex]
6. Simplify the equation:
[tex]\[ (x + 3)^2 + (y + 2)^2 - 9 - 4 = 3 \][/tex]
[tex]\[ (x + 3)^2 + (y + 2)^2 - 13 = 3 \][/tex]
[tex]\[ (x + 3)^2 + (y + 2)^2 = 16 \][/tex]
7. Write the equation in standard form:
[tex]\[ (x + 3)^2 + (y + 2)^2 = 4^2 \][/tex]
From this equation, the center of the circle [tex]\((h, k)\)[/tex] is [tex]\((-3, -2)\)[/tex] and the radius [tex]\( r \)[/tex] is 4.
So, the correct completion of the work is:
[tex]\[ (x + 3)^2 + (y + 2)^2 = 4^2 \][/tex]
Therefore, the center is [tex]\((-3, -2)\)[/tex] and the correct completion sentence is:
[tex]\[ \boxed{(x+3)^2+(y+2)^2=4^2,\ \text{so the center is}\ (-3, -2).} \][/tex]