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 1 comma negative 11 right parenthesis , radius=11

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

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

center=(−2,22), radius=41

Answer 1

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

Start with the given equation:

\[
x^2 - 2x + y^2 + 22y + 1 = 0
\]

Rearranging this gives:

\[
x^2 - 2x + y^2 + 22y = -1
\]

Now, complete the square for the \( x \) terms:

1. Take the coefficient of \( x \), which is \(-2\), halve it to get \(-1\), and then square it to get \(1\).
2. This gives us:
\[
x^2 - 2x = (x - 1)^2 - 1
\]

Next, complete the square for the \( y \) terms:

1. The coefficient of \( y \) is \(22\), halve it to get \(11\), and square it to get \(121\).
2. This gives us:
\[
y^2 + 22y = (y + 11)^2 - 121
\]

Now replace these into the equation:

\[
((x - 1)^2 - 1) + ((y + 11)^2 - 121) = -1
\]

Simplifying this results in:

\[
(x - 1)^2 + (y + 11)^2 - 122 = -1
\]

Add \(122\) to both sides:

\[
(x - 1)^2 + (y + 11)^2 = 121
\]

Now, we can identify the center and radius of the circle:

- The center \((h, k)\) is \((1, -11)\).
- The radius \(r\) is \(\sqrt{121} = 11\).

Therefore, the correct response is:

**center = (1, -11), radius = 11**