To identify the center of the circle from the equation \( x^2 + y^2 - 2x - 12y = 18 \), we will complete the square for both the \( x \) and \( y \) terms.
Step 1: Rearranging the equation
We can write the equation as:
\[ x^2 - 2x + y^2 - 12y = 18 \]
Step 2: Completing the square for \( x \)
Take the \( x \) terms: \( x^2 - 2x \).
- Take half of the coefficient of \( x \) (which is \(-2\)), square it: \[ \left(\frac{-2}{2}\right)^2 = 1 \]
- Add and subtract this square: \[ x^2 - 2x = (x - 1)^2 - 1 \]
Step 3: Completing the square for \( y \)
Take the \( y \) terms: \( y^2 - 12y \).
- Take half of the coefficient of \( y \) (which is \(-12\)), square it: \[ \left(\frac{-12}{2}\right)^2 = 36 \]
- Add and subtract this square: \[ y^2 - 12y = (y - 6)^2 - 36 \]
Step 4: Putting it all together
Now we can substitute the completed squares back into the equation:
\[ (x - 1)^2 - 1 + (y - 6)^2 - 36 = 18 \]
Combine constants:
\[ (x - 1)^2 + (y - 6)^2 - 37 = 18 \] \[ (x - 1)^2 + (y - 6)^2 = 55 \]
Step 5: Identify the center
From the standard form of the circle equation \((x - h)^2 + (y - k)^2 = r^2\), we see that the center \( (h, k) \) is given by:
- \( h = 1 \)
- \( k = 6 \)
Final Answer
Thus, the center of the circle is:
\[ \text{center: } (1, 6) \]