To find the center and radius of the circle described by the equation \(x^2 + y^2 + 2x - 8y + 16 = 0\), we need to rewrite the equation in the standard form of a circle, which is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
We can start by rearranging the given equation:
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Move the constant term to the other side of the equation:
\[ x^2 + y^2 + 2x - 8y + 16 = 0 \implies x^2 + y^2 + 2x - 8y = -16 \]
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Next, we will complete the square for the \(x\) and \(y\) terms.
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For \(x^2 + 2x\):
- Take half of the coefficient of \(x\) (which is 2), square it: \(\left(\frac{2}{2}\right)^2 = 1\).
- Therefore, \(x^2 + 2x\) can be rewritten as \((x + 1)^2 - 1\).
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For \(y^2 - 8y\):
- Take half of the coefficient of \(y\) (which is -8), square it: \(\left(\frac{-8}{2}\right)^2 = 16\).
- Therefore, \(y^2 - 8y\) can be rewritten as \((y - 4)^2 - 16\).
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Substitute these complete squares back into the equation:
\[ (x + 1)^2 - 1 + (y - 4)^2 - 16 = -16 \]
Simplifying gives:
\[ (x + 1)^2 + (y - 4)^2 - 17 = -16 \]
\[ (x + 1)^2 + (y - 4)^2 = 1 \]
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Now the equation is in standard form \((x + 1)^2 + (y - 4)^2 = 1\).
From this equation, we can identify:
- The center \((h, k)\) is \((-1, 4)\).
- The radius \(r\) is given by \(\sqrt{1} = 1\).
Thus, the center and radius of the circle are:
center: \((-1, 4)\), \(r = 1\).
The correct response is: center: \((-1, 4)\), r = 1.