Post a New Question

Pre Calc

posted by .

Complete the square to find the center and the radius of the circle:
3x^2+3y^2-12x+24y+15=0

  • Pre Calc -

    3x^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 -

    Thank you so much, i'm so stressed if its not too much trouble check out some of my other questions;)

  • Pre Calc -

    3x^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 -

    Of course 45รท3 = 15 and not 9 like I had.
    Silly me!

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question