Pre-Calc Question. Teacher is not helpful, please respond.?

Question

My pre-calc prof. is really hard and not too helpful at answering questions so here it goes: I need to find the center and radius of this equation 2x^2 - 8x + 2y^2 + 12y - 6 = 0. I understand how to perform the operation if the 2's were not present in front of the x and y^2, but with these extra numbers added in, I'm lost. I would really appreciate the help.

Answer 1

2x^2 - 8x + 2y^2 + 12y - 6 = 0

Use a process called completing the square. Add 6 to both sides.

2x^2 - 8x ............+ 2y^2 + 12y........... = 6

Factor the 2 out of each.

2(x^2 - 4x ..........) + 2 ( y^2 + 6y ............) = 6

Pull out the 4, divide it by 2, square the result.

4/2 = 2......square it...2^2 is 4. Add 4 in the parenthesis.  Do the same with the 6 in front of y.  

6/2 = 3....3^2 = 9.  Add 9 in the parenthesis with the y^2.

You are making perfect square trinomials, in other words...the trinomials can be factored into a binomial squared.

2(x^2 - 4x + 4) + 2 ( y^2 + 6y + 9) = 6 + 8 + 18

We add 8 and 18 to the right side to balance the equation because on the LEFT side, if you distribute the 2's, you get 2x^2 - 8x + 8 + 2y^2 + 12y + 18.  See how the 8 and 18 weren't in the original ?

2(x-2)^2 + 2(y+3)^2 = 32

General equation of a circle is (x-h)^2 + (y-k)^2 = r^2 so divide everything by 2.  The center is (h,k) and the radius is r.

(x-2)^2 + (y+3)^2 = 16

so the center is (2,-3) and the radius is 4.

Hope this helps and good luck this year !

Answer 2

once you divide all by 2 it will turn out;

x^2-4x + y^2+6y-3= 0  

Answer 3

Divide everything by 2.

2x^2 - 8x + 2y^2 + 12y - 6 = 0.

x^2 - 4x + y^2 + 6y - 3 = 0.

Complete the squares of both the x terms and the y terms.

Answer 4

2x^2-8x + 2y^2 + 12y - 6 = 0

x^2 - 4x + y^2 + 6y - 3 = 0

(x^2 - 4x + 4) + (y^2 + 6y + 9) = 16

(x-2)^2 + (y+3)^2 = 16

Center : (2,-3)

Radius = sqrt 16 = 4

Answer 5

I continue the previous solution(s)

x^2 - 4x + y^2 + 6y - 3 = 0.

is

x^2 - 4x + 4 + y^2 + 6y + 9 - 16 = 0

or

(x-2)^2 + (y+3)^2 = 16 = 4^2

The radius is 4.

Question to you: Where is the center of this circle?  

Answer 6

to get rid of the 2, divide by 2

2(x^2 - 4x + ?) + 2(y^2 + 6y + ??) = 6 + 2? + 2??

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

now you can do it

Answer 7

if you divide the entire equation by two you will get:

x^2-4x+y^2+6y-3=0

then you should be able to go from here to set up the equation of the circle