Expert: Sherman D. - 8/8/2006

Question
2(x^2 - 4x + 4 - 4) + 2(y^2 - 6y + 9 - 9) + 24 = 0

at this step, i don't understand where the +4 - 4 and the +9 - 9 came from.
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Followup To

Question -
Write the equation of the following circle in standard form. Find its center and radius.

2x² + 2y² - 8x - 12y + 24 = 0

Answer -
(x - h)^2 + (y - k)^2 = r^2

2x^2 + 2y^2 - 8x - 12y + 24 = 0
2x^2 - 8x + 2y^2 - 12y + 24 = 0
(2x^2 - 8x) + (2y^2 - 12y) + 24 = 0
2(x^2 - 4x) + 2(y^2 - 6y) + 24 = 0
2(x^2 - 4x + 4 - 4) + 2(y^2 - 6y + 9 - 9) + 24 = 0
2((x - 2)^2 - 4) + 2((y - 3)^2 - 9) + 24 = 0
2(x - 2)^2 - 8 + 2(y - 3)^2 - 18 + 24 = 0
2(x - 2)^2 + 2(y - 3)^2 - 2 = 0
2(x - 2)^2 + 2(y - 3)^2 = 2
(x - 2)^2 + (y - 3)^2 = 1

Center : (h,k)
Radius = r

(x - 2)^2 + (y - 3)^2 = 1
Center : (2,3)
Radius : 1

Answer
I thought you might ask that

Its using the completing the square method.

2(x^2 - 4x) + 2(y^2 - 6y) + 24 = 0

find half of 4x, square it, and add to both sides
find half of 6x, square it, and add to both sides

on the left, it will be negative, but on the right, you will be adding it.

Just take my word for it, if you follow the steps i have made, you can't go wrong.

If you have any trouble with other problems like this, you let me know.

or you can find me at answers.yahoo.com, trust me, you will recognize my name.