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Use the image to answer the question.

A circle is plotted on a coordinate plane.

Which equation is the standard form of the equation of a circle drawn on the coordinate plane, derived from the Pythagorean Theorem?

(1 point)
Responses

(x−6)2+(y−2)2=16
left parenthesis x minus 6 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 16

(x−2)2+(y−2)2=4
left parenthesis x minus 2 right parenthesis squared plus left parenthesis y minus 2 right parenthesis squared equals 4

(x+2)2+(y−6)2=4
left parenthesis x plus 2 right parenthesis squared plus left parenthesis y minus 6 right parenthesis squared equals 4

(x−2)2+(y+2)2=16

1 answer

The standard form of the equation of a circle is given by the equation:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \((h, k)\) is the center of the circle and \(r\) is the radius.

To determine which of the provided options corresponds to the standard form, we typically need information about the circle's center and radius. However, based on the general understanding of the format, we can analyze the options by extracting the center and radius squared from each equation.

  1. \((x−6)^2+(y−2)^2=16\)

    • Center: (6, 2), Radius: 4 (radius \(r = \sqrt{16}\))

  2. \((x−2)^2+(y−2)^2=4\)

    • Center: (2, 2), Radius: 2 (radius \(r = \sqrt{4}\))

  3. \((x+2)^2+(y−6)^2=4\)

    • Center: (-2, 6), Radius: 2 (radius \(r = \sqrt{4}\))

  4. \((x−2)^2+(y+2)^2=16\)

    • Center: (2, -2), Radius: 4 (radius \(r = \sqrt{16}\))

Without seeing the image to identify the center and radius, we can't definitively select an answer. However, if you know the center and radius, compare them with the options to find the correct standard form of the equation. Please provide the center and radius if you have that information, or describe the image for further assistance.