Question

Javier is drafting a circle on graph paper. Using the equation x2+y2−2x−12y=18
, complete the square to calculate where he would draw the center of the circle.(1 point)
center: (
,
)

Answer 1

To find the center of the circle defined by the equation \( x^2 + y^2 - 2x - 12y = 18 \), we first need to rearrange and complete the square for both the \( x \) and \( y \) terms.

1. **Reorganize the equation**:
\[
x^2 - 2x + y^2 - 12y = 18
\]

2. **Complete the square for \( x \)**:
- Take the coefficient of \( x \) which is -2, divide it by 2 to get -1, and square it to get 1.
- Add and subtract 1 in the equation:
\[
(x^2 - 2x + 1) + y^2 - 12y = 18 + 1
\]
- This simplifies to:
\[
(x - 1)^2 + y^2 - 12y = 19
\]

3. **Complete the square for \( y \)**:
- Take the coefficient of \( y \) which is -12, divide it by 2 to get -6, and square it to get 36.
- Add and subtract 36 in the equation:
\[
(x - 1)^2 + (y^2 - 12y + 36) = 19 + 36
\]
- This simplifies to:
\[
(x - 1)^2 + (y - 6)^2 = 55
\]

Now, we can see that the equation is in the standard form of a circle, \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

From our completed square results,
- \( h = 1 \)
- \( k = 6 \)

Thus, the center of the circle is:
\[
\text{center: } (1, 6)
\]