Find the area of the region enclosed by the graph of $x^2 + y^2 = 2x - 6y + 6$

  16π ≈ 50.27 square units

Step-by-step explanation:

The equation is that of a circle of radius 4. The area of a circle is given by ...

  A = πr²

  A = π(4²) = 16π ≈ 50.27 . . . . square units

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The equation can be put into the standard form for a circle to find its radius.

  (x^2 -2x) +(y^2 +6y) = 6 . . . . collect variable terms on the left

  (x^2 -2x +1) +(y^2 +6y +9) = 6 + 1 + 9 = 16 . . . . . complete the squares

  (x -1)^2 +(y +3)^2 = 4^2 . . . . . the radius is 4

Compare to the form ...

  (x -h)^2 +(y -k)^2 = r^2 . . . . circle of radius r centered at (h, k)

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.