Question

How do I write the equation of the circle in standard form, and determine the center and radius if it exists? for:

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

Answer 1

complete the squares if need be

Answer 2

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

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

Answer 3

2(x+2)^2 + 2(y-3)^2 = 8+8+18 = 34

Answer 4

the center is the opposites of the x and y modifiers...

C=(-2,3)

Answer 5

r^2 = 34

r = sqrt(34)

Answer 6

or do we divide it all by 2? first?

Answer 7

r=sqrt(17) prolly

Answer 8

Not easy, so do they both get divided by 2 then?

Answer 9

everything gets divided by 2...everything

Answer 10

Okay,

Answer 11

on a side note, we could divide everying by 34 to get:

(x+2)^2 (y-3)^2
------- + ------- = 1 ; which is the same equation
17 17

Answer 12

so I would stick with r = sqrt(17)

Answer 13

Okay, thank you

Answer 14

youre welcome :)