To find the center of the circle from the given equation \( x^2 + y - 2x - 12y = 18 \), we first need to rearrange and complete the square.
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Rearrange the equation: \[ x^2 - 2x + y - 12y = 18 \] Combine the \(y\) terms: \[ x^2 - 2x - 11y = 18 \]
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Rearranging gives: \[ x^2 - 2x + (-11y) = 18 \]
Next, we will complete the square for the \(x\) terms.
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For \(x^2 - 2x\):
- Take the coefficient of \(x\), which is \(-2\), divide by \(2\) (getting \(-1\)), and square it (\(1\)).
- Add and subtract \(1\): \[ (x^2 - 2x + 1) - 1 \] This gives us: \[ (x - 1)^2 - 1 \]
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Substitute back into the equation: \[ (x - 1)^2 - 1 - 11y = 18 \]
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Simplifying gives: \[ (x - 1)^2 - 11y = 19 \]
Now, solve for \(y\): \[ -11y = 19 - (x - 1)^2 \] \[ y = -\frac{1}{11} (19 - (x - 1)^2) \]
This means we need to separately isolate the \(y\) terms correctly for the completion of the square correctly. So let’s revise and rewrite the equation again properly focusing on \(y\):
Let's break down: From \(x^2 - 2x + y - 12y = 18\) it becomes quick to rearrange into:
- Move all others to one side: \(x^2 - 2x + y(1-12) = 18\) but we are complexly noting that we need to complete to circles.
So we think in circle form \( (x - h)^2 + (y - k)^2 = r^2 \).
From: To correctly observe \(x^2 - 2x + 1 -1 + ...\)
We realize clearly:
\[(x - 1)^2 + (-11)(y - \frac{12}{2}) =19\]
Thus clearly leads to the center definition \( (1,\frac{12}{11})\).
Conclusion:
The center of the circle is at the coordinates: \( \text{center: } (1, \frac{12}{11})\).