To find the center of the circle represented by the equation \( x^2 + y^2 - 2x - 12y = 18 \), we need to rewrite it in standard form by completing the square for both \( x \) and \( y \).
Step 1: Rearrange the equation
First, we can rewrite the equation by moving the constant to the other side: \[ x^2 + y^2 - 2x - 12y - 18 = 0 \]
Step 2: Group the \( x \) and \( y \) terms
Group the \( x \) terms and the \( y \) terms: \[ (x^2 - 2x) + (y^2 - 12y) = 18 \]
Step 3: Complete the square for \( x \)
For the \( x \) terms, take the coefficient of \( x \), which is \(-2\), halve it to get \(-1\), and square it to get \(1\): \[ x^2 - 2x = (x - 1)^2 - 1 \]
Step 4: Complete the square for \( y \)
For the \( y \) terms, take the coefficient of \( y \), which is \(-12\), halve it to get \(-6\), and square it to get \(36\): \[ y^2 - 12y = (y - 6)^2 - 36 \]
Step 5: Rewrite the equation
Plug the completed squares back into the equation: \[ ((x - 1)^2 - 1) + ((y - 6)^2 - 36) = 18 \] Now, distribute and combine: \[ (x - 1)^2 + (y - 6)^2 - 1 - 36 = 18 \] \[ (x - 1)^2 + (y - 6)^2 - 37 = 18 \] Now add \(37\) to both sides: \[ (x - 1)^2 + (y - 6)^2 = 55 \]
Step 6: Determine the center
From the equation \((x - 1)^2 + (y - 6)^2 = 55\), we can see that the center of the circle is at the point \((1, 6)\).
Final Answer
The center of the circle is: \[ \text{center: } (1, 6) \]