Question

Regina is trying to sketch a circle on graph paper with the equation x^2 + y^2 - 2x + 22y + 1 = 0. Calculate where she would need to graph the center and radius of the circle by completing the square.

a. center = (-1, 11), radius = 11
b. center = (1, -11), radius = 11
c. center = (-2 , 22), radius = 41
d. center = (-22 , 2), radius = 41

Answer 1

To find the center and radius of the circle given by the equation \( x^2 + y^2 - 2x + 22y + 1 = 0 \), we can complete the square.

1. Start by rearranging the equation to isolate the constant on the right side:
\[
x^2 - 2x + y^2 + 22y + 1 = 0 \implies x^2 - 2x + y^2 + 22y = -1
\]

2. Now, complete the square for \( x \) and \( y \).

**For \( x^2 - 2x \):**
- Take half of the coefficient of \( x \), square it:
\[
\left(\frac{-2}{2}\right)^2 = 1
\]
- Rewrite it as:
\[
(x - 1)^2 - 1
\]

**For \( y^2 + 22y \):**
- Take half of the coefficient of \( y \), square it:
\[
\left(\frac{22}{2}\right)^2 = 121
\]
- Rewrite it as:
\[
(y + 11)^2 - 121
\]

3. Substitute these completed squares back into the equation:
\[
(x - 1)^2 - 1 + (y + 11)^2 - 121 = -1
\]
Simplifying gives:
\[
(x - 1)^2 + (y + 11)^2 - 122 = -1
\]
\[
(x - 1)^2 + (y + 11)^2 = 121
\]

4. From the equation \( (x - 1)^2 + (y + 11)^2 = 121 \), we can identify the center and radius:
- The center (h, k) is \( (1, -11) \)
- The radius \( r \) is \( \sqrt{121} = 11 \)

Thus, the center of the circle is \( (1, -11) \) and the radius is \( 11 \).

The correct option is:
**b. center = (1, -11), radius = 11**.