To find the center of the circle defined by the equation \(x^2 + y^2 - 6x - 12y + 5 = 0\), we will complete the square for both \(x\) and \(y\).
-
Start with the equation:
\[ x^2 + y^2 - 6x - 12y + 5 = 0 \]
-
Rearrange the equation to group the \(x\) and \(y\) terms:
\[ x^2 - 6x + y^2 - 12y + 5 = 0 \]
-
Move the constant to the other side:
\[ x^2 - 6x + y^2 - 12y = -5 \]
-
Complete the square for the \(x\) terms:
For \(x^2 - 6x\):
- Take half of \(-6\) to get \(-3\) and square it to get \(9\).
- Thus, \(x^2 - 6x\) becomes \((x - 3)^2 - 9\).
-
Complete the square for the \(y\) terms:
For \(y^2 - 12y\):
- Take half of \(-12\) to get \(-6\) and square it to get \(36\).
- Thus, \(y^2 - 12y\) becomes \((y - 6)^2 - 36\).
-
Substitute the completed squares back into the equation:
\[ (x - 3)^2 - 9 + (y - 6)^2 - 36 = -5 \]
Simplify this:
\[ (x - 3)^2 + (y - 6)^2 - 45 = -5 \]
Which becomes:
\[ (x - 3)^2 + (y - 6)^2 = 40 \]
-
From the standard form of the equation of a circle \((x - h)^2 + (y - k)^2 = r^2\), we can identify the center of the circle \( (h, k) \):
- Here, \(h = 3\) and \(k = 6\).
Thus, the center of the circle is at:
\[ \boxed{(3, 6)} \]