To find the center of the circle defined by the equation \( x^2 + y^2 - 2x - 12y = 18 \), we first need to rearrange and complete the square for both the \( x \) and \( y \) terms.
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Reorganize the equation: \[ x^2 - 2x + y^2 - 12y = 18 \]
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Complete the square for \( x \):
- Take the coefficient of \( x \) which is -2, divide it by 2 to get -1, and square it to get 1.
- Add and subtract 1 in the equation: \[ (x^2 - 2x + 1) + y^2 - 12y = 18 + 1 \]
- This simplifies to: \[ (x - 1)^2 + y^2 - 12y = 19 \]
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Complete the square for \( y \):
- Take the coefficient of \( y \) which is -12, divide it by 2 to get -6, and square it to get 36.
- Add and subtract 36 in the equation: \[ (x - 1)^2 + (y^2 - 12y + 36) = 19 + 36 \]
- This simplifies to: \[ (x - 1)^2 + (y - 6)^2 = 55 \]
Now, we can see that the equation is in the standard form of a circle, \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
From our completed square results,
- \( h = 1 \)
- \( k = 6 \)
Thus, the center of the circle is: \[ \text{center: } (1, 6) \]
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