Determine center point and radius in the equation x^2+y^2-6y+1=0

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Step-by-step explanation:

The equation can be put into standard form by "completing the square" for each variable. The x-term is already a square, so we need do this only for the y-term.

  x^2 +(y^2 -6y) = -1

  x^2 +(y^2 -6y +9) = -1 +9 . . . . . . add (-6/2)^2 to complete the square

  x^2 +(y -3)^2 = 8 . . . . standard form equation

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Comparing the above equation to the form:

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

we see that (h, k) = (0, 3) and r^2 = 8 ⇒ r = 2√2.

The center point is (0, 3); the radius is 2√2.

step-by-step explanation: