To convert the given equation of the circle from general form to standard form, we start with the equation:
\[ x^2 + y^2 + 6x - 4y + 4 = 0. \]
First, we reorganize it:
\[ x^2 + 6x + y^2 - 4y + 4 = 0. \]
Next, we move the constant term to the other side:
\[ x^2 + 6x + y^2 - 4y = -4. \]
Now we will complete the square for both \(x\) and \(y\).
For \(x^2 + 6x\):
- Take the coefficient of \(x\), which is 6, divide by 2 to get 3, and square it to get 9.
- Thus, we can write: \[ x^2 + 6x = (x + 3)^2 - 9. \]
For \(y^2 - 4y\):
- Take the coefficient of \(y\), which is -4, divide by 2 to get -2, and square it to get 4.
- Thus, we can write: \[ y^2 - 4y = (y - 2)^2 - 4. \]
Substituting these back into the equation:
\[(x + 3)^2 - 9 + (y - 2)^2 - 4 = -4.\]
Simplifying this gives:
\[(x + 3)^2 + (y - 2)^2 - 13 = -4,\]
which simplifies to:
\[(x + 3)^2 + (y - 2)^2 = 9.\]
Now the equation is in standard form:
\[(x + 3)^2 + (y - 2)^2 = 9.\]
Thus, the correct response is:
left-parenthesis x plus 3 right-parenthesis squared plus left-parenthesis y minus 2 right-parenthesis squared equals 9.