Question

Calculate the center of the circle by completing the square of the equation x2+y2−6x−12y+5=0
.(1 point)
(
,
)

Answer 1

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 \(y\).

1. Start with the equation:

\[
x^2 + y^2 - 6x - 12y + 5 = 0
\]

2. Rearrange the equation to group the \(x\) and \(y\) terms:

\[
x^2 - 6x + y^2 - 12y + 5 = 0
\]

3. Move the constant to the other side:

\[
x^2 - 6x + y^2 - 12y = -5
\]

4. 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 \((x - 3)^2 - 9\).

5. 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 \((y - 6)^2 - 36\).

6. 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:

\[
(x - 3)^2 + (y - 6)^2 = 40
\]

7. 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:

\[
\boxed{(3, 6)}
\]