Question
The equation of a circle in general form x^2 + y^2 + 6x - 4y + 4 = 0 . Write the equation in standard form. (1 point)
Answers
GPT-4o mini
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:
\((x + 3)^2 + (y - 2)^2 = 1\)
\[
x^2 + y^2 + 6x - 4y + 4 = 0
\]
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:
\((x + 3)^2 + (y - 2)^2 = 1\)