Problem

For \((x~-~2)^2~+~(y~+~1)^2~=~9\), give the circle's center, radius squared, positive radius, \(x\text{-}\text{extent}\), and \(y\text{-}\text{extent}\).

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What this problem is really about

Match the equation directly to circle standard form before computing anything. Read the center with signs opposite the binomials, take the nonnegative square root of the right side for the radius, and then move one radius left, right, down, and up from the center. Those four anchor points determine the horizontal and vertical extents.

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The center is (2, -1), the radius squared is 9, the radius is 9, the x-extent is [-7, 11], and the y-extent is [-10, 8].

The center is (2, -1), the radius squared is 36, the radius is 6, the x-extent is [-4, 8], and the y-extent is [-7, 5].

The center is (-2, 1), the radius squared is 9, the radius is 3, the x-extent is [-5, 1], and the y-extent is [-2, 4].

The center is (2, -1), the radius squared is 9, the radius is 3, the x-extent is [-1, 5], and the y-extent is [-4, 2].

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For \((x~-~2)^2~+~(y~+~1)^2~=~9\), give the circle's center, radius squared, positive radius, \(x\text{-}\text{extent}\), and \(y\text{-}\text{extent}\).

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