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.

geometry: triangles

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call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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