To find the center of the circle represented by the equation \( x^2 + y^2 - 2x - 12y = 18 \), we need to complete the square for both \( x \) and \( y \).
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Rearranging the Equation: We start with the original equation: \[ x^2 + y^2 - 2x - 12y = 18 \] We can move the constant to the right side: \[ x^2 - 2x + y^2 - 12y = 18 \]
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Completing the Square for \( x \): Take the \( x \) terms: \[ x^2 - 2x \] To complete the square, take half of the coefficient of \( x \) (which is \(-2\)), square it, and add it: \[ \left(-\frac{2}{2}\right)^2 = 1 \] When you add 1, you need to also subtract 1 to keep the equation balanced. Thus: \[ x^2 - 2x = (x - 1)^2 - 1 \]
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Completing the Square for \( y \): Now take the \( y \) terms: \[ y^2 - 12y \] Again, take half of \(-12\) and square it: \[ \left(-\frac{12}{2}\right)^2 = 36 \] Similarly, add and subtract 36: \[ y^2 - 12y = (y - 6)^2 - 36 \]
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Substituting Back into the Equation: We can now rewrite the equation with completed squares: \[ (x - 1)^2 - 1 + (y - 6)^2 - 36 = 18 \] Combine the constants: \[ (x - 1)^2 + (y - 6)^2 - 37 = 18 \] \[ (x - 1)^2 + (y - 6)^2 = 18 + 37 \] \[ (x - 1)^2 + (y - 6)^2 = 55 \]
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Finding the Center: From the standard circle equation form \( (x - h)^2 + (y - k)^2 = r^2 \), we can see:
- The center \((h, k)\) is \((1, 6)\)
- The radius \(r\) is \(\sqrt{55}\) (not necessary for the answer, but good to note)
Thus, Javier should draw the center of the circle at the point \((1, 6)\).