To find the center and radius of the circle given by the equation \(x^2 + y^2 - 2x + 22y + 1 = 0\), we need to complete the square for both the \(x\) and \(y\) terms.
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Start by rearranging the equation: \[ x^2 - 2x + y^2 + 22y + 1 = 0 \] Now, move the constant term to the other side: \[ x^2 - 2x + y^2 + 22y = -1 \]
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Complete the square for \(x^2 - 2x\): \[ x^2 - 2x \rightarrow (x - 1)^2 - 1 \]
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Complete the square for \(y^2 + 22y\): \[ y^2 + 22y \rightarrow (y + 11)^2 - 121 \]
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Substitute the completed squares back into the equation: \[ (x - 1)^2 - 1 + (y + 11)^2 - 121 = -1 \] Simplifying gives: \[ (x - 1)^2 + (y + 11)^2 - 122 = -1 \] \[ (x - 1)^2 + (y + 11)^2 = 121 \]
Now we can identify the center and the radius of the circle:
- The center \((h, k)\) is \((1, -11)\).
- The radius \(r\) is \(\sqrt{121} = 11\).
So, the center is \((1, -11)\) and the radius is \(11\).
The correct response is: center=(1,−11), radius=11.