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.

Answer 1

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

1. **Rearranging the Equation:**
We start with the original equation:
\[
x^2 + y^2 - 2x - 12y = 18
\]
We can move the constant to the right side:
\[
x^2 - 2x + y^2 - 12y = 18
\]

2. **Completing the Square for \( x \):**
Take the \( x \) terms:
\[
x^2 - 2x
\]
To complete the square, take half of the coefficient of \( x \) (which is \(-2\)), square it, and add it:
\[
\left(-\frac{2}{2}\right)^2 = 1
\]
When you add 1, you need to also subtract 1 to keep the equation balanced. Thus:
\[
x^2 - 2x = (x - 1)^2 - 1
\]

3. **Completing the Square for \( y \):**
Now take the \( y \) terms:
\[
y^2 - 12y
\]
Again, take half of \(-12\) and square it:
\[
\left(-\frac{12}{2}\right)^2 = 36
\]
Similarly, add and subtract 36:
\[
y^2 - 12y = (y - 6)^2 - 36
\]

4. **Substituting Back into the Equation:**
We can now rewrite the equation with completed squares:
\[
(x - 1)^2 - 1 + (y - 6)^2 - 36 = 18
\]
Combine the constants:
\[
(x - 1)^2 + (y - 6)^2 - 37 = 18
\]
\[
(x - 1)^2 + (y - 6)^2 = 18 + 37
\]
\[
(x - 1)^2 + (y - 6)^2 = 55
\]

5. **Finding the Center:**
From the standard circle equation form \( (x - h)^2 + (y - k)^2 = r^2 \), we can see:
- The center \((h, k)\) is \((1, 6)\)
- The radius \(r\) is \(\sqrt{55}\) (not necessary for the answer, but good to note)

Thus, Javier should draw the center of the circle at the point \((1, 6)\).