The equation of a circle in general form x^2 + y^2 + 6x - 4y + 4 = 0 . Write the equation in standard form. (1 point)

Answer 1

To convert the equation of the circle from general form to standard form, we start with the given equation:

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

First, rearrange the equation by moving the constant to the right side:

\[ x^2 + y^2 + 6x - 4y = -4 \]

Next, we need to complete the square for the \(x\) and \(y\) terms.

Completing the square for \(x\):

The \(x\) terms are \(x^2 + 6x\). To complete the square:

1. Take half of the coefficient of \(x\) (which is 6), square it, and add/subtract it.
   • Half of 6 is 3, and \(3^2 = 9\).

Thus,

\[ x^2 + 6x = (x + 3)^2 - 9 \]

Completing the square for \(y\):

The \(y\) terms are \(-4y\). To complete the square:

1. Take half of the coefficient of \(y\) (which is -4), square it, and add/subtract it.
   • Half of -4 is -2, and \((-2)^2 = 4\).

Thus,

\[ -4y = -(y^2 - 4y) = -((y - 2)^2 - 4) = -(y - 2)^2 + 4 \]

Putting it together:

Substituting these completed squares back into the equation:

\[ (x + 3)^2 - 9 - (y - 2)^2 + 4 = -4 \]

This simplifies as follows:

\[ (x + 3)^2 - (y - 2)^2 - 5 = -4 \]

Now combine the constants on the right:

\[ (x + 3)^2 - (y - 2)^2 = -4 + 5 \]

Which simplifies to:

\[ (x + 3)^2 - (y - 2)^2 = 1 \]

Finally, to convert to standard form, we write it as:

\[ (x + 3)^2 + (y - 2)^2 = 1 \]

Thus, the equation of the circle in standard form is:

\((x + 3)^2 + (y - 2)^2 = 1\)