To find the center and radius of the circle represented by the equation \(x^2 + y^2 - 2x + 6y + 6 = 0\), we can rewrite the equation in the standard form of a circle.
The standard form of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
Let's start with the original equation:
\[ x^2 + y^2 - 2x + 6y + 6 = 0 \]
We can rearrange it to:
\[ x^2 - 2x + y^2 + 6y + 6 = 0 \]
Next, we will complete the square for both \(x\) and \(y\).
Completing the square for \(x\):
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Take the coefficient of \(x\) (which is \(-2\)), divide by 2 (giving \(-1\)), and square it (giving \(1\)):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Completing the square for \(y\):
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Take the coefficient of \(y\) (which is \(6\)), divide by 2 (giving \(3\)), and square it (giving \(9\)):
\[ y^2 + 6y = (y + 3)^2 - 9 \]
Substitute back into the equation:
Now substituting these back into the equation gives us:
\[ (x - 1)^2 - 1 + (y + 3)^2 - 9 + 6 = 0 \]
Simplifying this:
\[ (x - 1)^2 + (y + 3)^2 - 4 = 0 \] \[ (x - 1)^2 + (y + 3)^2 = 4 \]
Now we can identify the center and radius from this standard form:
- The center \((h, k)\) is \((1, -3)\).
- The radius \(r\) is \(\sqrt{4} = 2\).
Thus, the center of the circle is \((1, -3)\) and the radius is \(2\).
The correct response is:
center: (1, -3) radius: 2