Question

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

Answer 1

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)

Answer 2

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

Answer 3

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.

Answer 4

Of course 45÷3 = 15 and not 9 like I had.
Silly me!