Asked by Jaz
The graph of a circle is shown on a coordinate grid with center at left-parenthesis negative 2 comma 2 right-parenthesis passing through left-parenthesis 1 comma 0 right-parenthesis and left-parenthesis negative 5 comma 0 right-parenthesis.
Question
Use the image to answer the question.
Write the standard form of the equation of the circle shown.
(1 point)
Responses
left parenthesis x plus 2 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals Start Root 13 End Root
Image with alt text: left parenthesis x plus 2 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals Start Root 13 End Root
left parenthesis x minus 2 right parenthesis squared plus left parenthesis y plus 2 right parenthesis squared equals 13
Image with alt text: left parenthesis x minus 2 right parenthesis squared plus left parenthesis y plus 2 right parenthesis squared equals 13
left parenthesis x plus 2 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 13
Image with alt text: left parenthesis x plus 2 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 13
left parenthesis x minus 2 right parenthesis squared plus left parenthesis y plus 2 right parenthesis squared equals Start Root 13 End Root
Image with alt text: left parenthesis x minus 2 right parenthesis squared plus left parenthesis y plus 2 right parenthesis squared equals Start Root 13 End Root
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Question
Use the image to answer the question.
Write the standard form of the equation of the circle shown.
(1 point)
Responses
left parenthesis x plus 2 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals Start Root 13 End Root
Image with alt text: left parenthesis x plus 2 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals Start Root 13 End Root
left parenthesis x minus 2 right parenthesis squared plus left parenthesis y plus 2 right parenthesis squared equals 13
Image with alt text: left parenthesis x minus 2 right parenthesis squared plus left parenthesis y plus 2 right parenthesis squared equals 13
left parenthesis x plus 2 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 13
Image with alt text: left parenthesis x plus 2 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 13
left parenthesis x minus 2 right parenthesis squared plus left parenthesis y plus 2 right parenthesis squared equals Start Root 13 End Root
Image with alt text: left parenthesis x minus 2 right parenthesis squared plus left parenthesis y plus 2 right parenthesis squared equals Start Root 13 End Root
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Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find the standard form of the equation of the circle, we use the general formula for a circle:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \((h, k)\) is the center of the circle and \(r\) is the radius.
Given:
- Center of the circle \( (h, k) = (-2, 2) \)
- Points on the circle \( (1, 0) \) and \( (-5, 0) \)
First, we find the radius \(r\). We can calculate the distance between the center and one of the points on the circle, for example, from \((-2, 2)\) to \((1, 0)\):
\[
r = \sqrt{(1 - (-2))^2 + (0 - 2)^2} = \sqrt{(1 + 2)^2 + (0 - 2)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]
Now that we have the center \((-2, 2)\) and the radius \(r = \sqrt{13}\), we can substitute these values into the circle equation:
\[
(x + 2)^2 + (y - 2)^2 = (\sqrt{13})^2
\]
This simplifies to:
\[
(x + 2)^2 + (y - 2)^2 = 13
\]
Thus, the standard form of the equation of the circle is:
\[
\boxed{(x + 2)^2 + (y - 2)^2 = 13}
\]
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \((h, k)\) is the center of the circle and \(r\) is the radius.
Given:
- Center of the circle \( (h, k) = (-2, 2) \)
- Points on the circle \( (1, 0) \) and \( (-5, 0) \)
First, we find the radius \(r\). We can calculate the distance between the center and one of the points on the circle, for example, from \((-2, 2)\) to \((1, 0)\):
\[
r = \sqrt{(1 - (-2))^2 + (0 - 2)^2} = \sqrt{(1 + 2)^2 + (0 - 2)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]
Now that we have the center \((-2, 2)\) and the radius \(r = \sqrt{13}\), we can substitute these values into the circle equation:
\[
(x + 2)^2 + (y - 2)^2 = (\sqrt{13})^2
\]
This simplifies to:
\[
(x + 2)^2 + (y - 2)^2 = 13
\]
Thus, the standard form of the equation of the circle is:
\[
\boxed{(x + 2)^2 + (y - 2)^2 = 13}
\]
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