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August 28, 2014

August 28, 2014

Posted by **Connor** on Tuesday, August 31, 2010 at 7:31pm.

3x^2+3y^2-12x+24y+15=0

- Pre Calc -
**Reiny**, Tuesday, August 31, 2010 at 7:55pm3x^2+3y^2-12x+24y+15=0

3(x^2 - 4x + ...) + 3(y^2 + 8y + ...) = -15

3(x^2 - 4x + 4) + 3(y^2 + 8y + 16) = -15 + 12 + 48

3(x-2)^2 + 3(y+4)^2 = 45

(x-2)^2 + 3(y+4)^2 = 9

take it from there.

(I suppose I could have divided each term by 3 at the start)

- Pre Calc -
**Connor**, Tuesday, August 31, 2010 at 7:59pmThank you so much, i'm so stressed if its not too much trouble check out some of my other questions;)

- Pre Calc -
**Henry**, Tuesday, August 31, 2010 at 8:43pm3x^2 + 3y^2 - 12x + 24y + 15 = 0.

Divide both sides of Eq by 3 to reduce

the coefficients of x^2 and y^2 to one:

x^2 + y^2 - 4x + 8y + 5 = 0

Complete the squares:

x^2 -4x + (-4/2)^2 + y^2 + 8y +(8/2)^2,

The terms that were added to complete

the square should be added to rt side

also:

x^2 -4x + 4 + y^2 +8y +16 = 4 +16 - 5,

Write the perfect squares as binomials

squared:

(x - 2)^2 + (x + 4)^2 = 15,

(x - h)^2 + (y - k)^2 = r^2 = STD Form.

C(h , k) = C(2 , -4), r^2 = 15, r =

sqrt(15) = 3.89.

- Thanks Henry - Pre Calc -
**Reiny**, Tuesday, August 31, 2010 at 9:15pmOf course 45÷3 = 15 and not 9 like I had.

Silly me!

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