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

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

amistre64 · Answer

complete the squares if need be

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

the center is the opposites of the x and y modifiers...

C=(-2,3)

amistre64 · Answer

r^2 = 34

r = sqrt(34)

amistre64 · Answer

or do we divide it all by 2? first?

amistre64 · Answer

r=sqrt(17) prolly

anonymous · Answer

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

amistre64 · Answer

everything gets divided by 2...everything

anonymous · Answer

Okay,

amistre64 · Answer

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

amistre64 · Answer

so I would stick with r = sqrt(17)

anonymous · Answer

Okay, thank you

amistre64 · Answer

youre welcome :)