Question

Regina is trying to sketch a circle on graph paper with the equation x2+y2−2x+22y+1=0. Calculate where she would need to graph the center and radius of the circle by completing the square.(1 point)
Responses

center=(−1,11), radius=11
center= left parenthesis negative 1 comma 11 right parenthesis , radius=11

center=(1,−11), radius=11
center= left parenthesis 1 comma negative 11 right parenthesis , radius=11

center=(−2,22), radius=41
center= left parenthesis negative 2 comma 22 right parenthesis , radius=41

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 will complete the square for both the \(x\) and \(y\) terms.

1. Start with the original equation:
\[
x^2 + y^2 - 2x + 22y + 1 = 0
\]

2. Rearrange it to group the \(x\) and \(y\) terms:
\[
(x^2 - 2x) + (y^2 + 22y) + 1 = 0
\]

3. Now, complete the square for \(x\):
- The term \(x^2 - 2x\) can be completed as follows:
\[
x^2 - 2x = (x - 1)^2 - 1
\]

4. Next, complete the square for \(y\):
- The term \(y^2 + 22y\) can be completed as follows:
\[
y^2 + 22y = (y + 11)^2 - 121
\]

5. Substitute back into the equation:
\[
((x - 1)^2 - 1) + ((y + 11)^2 - 121) + 1 = 0
\]
Simplifying this:
\[
(x - 1)^2 + (y + 11)^2 - 121 = 0
\]
\[
(x - 1)^2 + (y + 11)^2 = 121
\]

6. This 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 this, we can determine:
- Center: \((1, -11)\)
- Radius: \(\sqrt{121} = 11\)

Thus, the answer is:
**center=(1,−11), radius=11**.