To find the center and radius of the circle given by the equation:
\[ x^2 + y^2 + 8x - 6y - 11 = 0, \]
we first rearrange it into the standard form of a circle's equation:
\[ (x - h)^2 + (y - k)^2 = r^2, \]
where \((h, k)\) is the center and \(r\) is the radius.
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Group the \(x\) and \(y\) terms: \[ x^2 + 8x + y^2 - 6y = 11. \]
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Complete the square for the \(x\) terms: \[ x^2 + 8x \rightarrow (x + 4)^2 - 16. \]
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Complete the square for the \(y\) terms: \[ y^2 - 6y \rightarrow (y - 3)^2 - 9. \]
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Substituting back into the equation: \[ (x + 4)^2 - 16 + (y - 3)^2 - 9 = 11. \] Simplifying this gives: \[ (x + 4)^2 + (y - 3)^2 - 25 = 11. \] \[ (x + 4)^2 + (y - 3)^2 = 36. \]
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Identify the center and radius:
- The standard form indicates that the center \((h, k)\) is \((-4, 3)\) and the radius \(r\) is \(\sqrt{36} = 6\).
Therefore, the final answer is:
Center: (-4, 3), Radius: 6.