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 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: \( x^2 - 2x \).

1. Take half of the coefficient of \( x \) (which is \(-2\)), square it:
\[
\left(\frac{-2}{2}\right)^2 = 1
\]
2. Add and subtract this square:
\[
x^2 - 2x = (x - 1)^2 - 1
\]

### Step 3: Completing the square for \( y \)

Take the \( y \) terms: \( y^2 - 12y \).

1. Take half of the coefficient of \( y \) (which is \(-12\)), square it:
\[
\left(\frac{-12}{2}\right)^2 = 36
\]
2. 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:

\[
\text{center: } (1, 6)
\]