Question

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Using Equations of Circles Quick Check
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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=(−2,22)
, radius=41
center= left parenthesis negative 2 comma 22 right parenthesis , radius=41

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

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

center=(1,−11)
, radius=11

Answer 1

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

1. Start by rearranging the equation:
\[
x^2 - 2x + y^2 + 22y + 1 = 0
\]
Now, move the constant term to the other side:
\[
x^2 - 2x + y^2 + 22y = -1
\]

2. Complete the square for \(x^2 - 2x\):
\[
x^2 - 2x \rightarrow (x - 1)^2 - 1
\]

3. Complete the square for \(y^2 + 22y\):
\[
y^2 + 22y \rightarrow (y + 11)^2 - 121
\]

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

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

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

So, the center is \((1, -11)\) and the radius is \(11\).

The correct response is:
**center=(1,−11), radius=11**.